Epic 面试专题
由于3月5号要进行Epic公司的Online Assessment，准备刷些面经做准备，毕竟这是我的第一次比较重要的面试。。
这是我准备Epic技术面试的小专栏。。。希望面试的时候能做够30题+吧，这样做OA的时候会轻松一点。
目前完成：29题
/*************************************************我是分割线*******************************************************/
虽然准备的比较充分了，但是还是悲剧了。。
原因可能有以下几点：
1，确实招人快招满了
2，电话面试的设计题答的一坨屎。。另外三哥哥的口音仍然听不懂
3，题做的还是不够多
不管怎样，仍然努力刷题+提升自己背景双管齐下吧。。。提升背景是为了拿到更多面试，刷题是为了在面试中发挥的更优秀。Epic是我半年北美求职之路的开始。希望在今年十月毕业的时候能有好的offer吧！
/*********************************************************************************************************************/
Epic 01 Colorful Number
本题题目要求如下：
/* Description: Determine whether a number is colorful or not.
* 263 is a colorful number because (2,6,3,2x6,6x3,2x3x6) are all different
* whereas 236 is not because (2,3,6,2x3,3x6,2x3x6) have 6 twice.
* So take all consecutive subsets of digits, take their product
* and ensure all the products are different */我对这个的算法就是两次循环
第一次：顺序加入2,6,3
每次都检查
第二次：
检查2 * 6， 2 * 6 * 3，
检查6 * 3
实现代码如下：
# include
# include
# include
# include
using namespace std;
bool color(int);
int main(int argc, char** argv) {
// cout << color(263) << endl;
cout << color(236) << endl;
// cout << color(40) << endl;
// cout << color(41) << endl;
// cout << color(2357496) << endl;
// cout << color(2634) << endl;
// cout << color(0) << endl;
// cout << color(9) << endl;
return 0;
}
bool color(int num) {
if (num < 0)
return false;
set s;
string str = to_string(num);
int color = str[0] - '0';
s.insert(color);
for (int i = 1; i < str.length(); ++i) {
int n = str[i] - '0';
cout << "check: " << n << endl;
if (s.find(n) != s.end())
return false;
else
s.insert(n);
}
for (int i = 0; i < str.length(); ++i) {
color = str[i] - '0';
for (int j = i + 1; j < str.length(); ++j) {
int n = str[j] - '0';
color *= n;
cout << "check: " << color << endl;
if (s.find(color) != s.end())
return false;
else
s.insert(color);
}
}
return true;
} Epic 02 Well ordered password
本题题目要求如下：
/* Description: Find all the possible passwords, given the length of the password and that
* it is a well ordered number (159 is well-ordered as 1<5<9) */举个简单的例子：
下面是我画的一张图：
下面的就是假设密码是0-4然后密码只有2位的解。
由于第一层时，还没有数字，所以剩余2位，则最大数字是5-2=3.最小数字是parent + 1，但是没有parent，所以是0
第二层，最小可取的是parent + 1，最大是 5 - 1 = 4
这个基本就跟recursive解dfs一样，代码如下：
# include
# include
using namespace std;
void well_ordered(int, int, int);
vector res;
int main(int argc, char** argv) {
well_ordered(0, 0, 4);
for (int i = 0; i < res.size(); ++i)
cout << res[i] << endl;
}
void well_ordered(int num, int start, int remain) {
if (remain == 0)
res.push_back(num);
else {
for (int i = start + 1; i <= (10 - remain); ++i) {
well_ordered(num * 10 + i, i, remain - 1);
}
}
} Epic 03 Spiral order
题目要求如下：
/* Description: If an N X N matrix is given, print it in spiral order.
* Example: Below is 5 X 5 matrix
* i l o v e
* d i n t e
* n i e e p
* a v w r i
* m a x e c
* Print in spiral order. Output is iloveepicexamandinterview */本题曾经在leetcode中做过，我的博客里面也有这道题的分析，这里就不重复了，直接上传代码：
# include
# include
using namespace std;
vector spiral_order(vector>);
int main(int argc, char** argv) {
vector> str = {
{'i', 'l', 'o', 'v', 'e'},
{'d', 'i', 'n', 't', 'e'},
{'n', 'e', 'w', 'e', 'p'},
{'a', 'i', 'v', 'r', 'i'},
{'m', 'a', 'x', 'e', 'c'}
};
vector res = spiral_order(str);
for(int i = 0; i < res.size(); ++i)
cout << res[i];
cout << endl;
return 0;
}
vector spiral_order(vector> str) {
vector res;
int col_start = 0;
int col_end = 4;
int row_start = 0;
int row_end = 4;
while (true) {
/* go right */
for (int i = col_start; i <= col_end; ++i)
res.push_back(str[row_start][i]);
row_start += 1;
if (res.size() == 25)
break;
/* go down */
for (int j = row_start; j <= row_end; ++j)
res.push_back(str[j][col_end]);
col_end -= 1;
if (res.size() == 25)
break;
/* go left */
for (int i = col_end; i >= col_start; --i)
res.push_back(str[row_end][i]);
row_end -= 1;
if (res.size() == 25)
break;
/* go up */
for (int j = row_end; j >= row_start; --j)
res.push_back(str[j][col_start]);
col_start += 1;
if (res.size() == 25)
break;
}
return res;
} Epic 04 Allowed password
本题要求如下：
/* Description: There is a security keypad at the entrance of a building.
* It has 9 numbers 1 - 9 in a 3x3 matrix format.
* 1 2 3
* 4 5 6
* 7 8 9
* The security has decided to allow one digit error for a person but that digit
* should be horizontal or vertical. Example: for 5 the user is allowed to enter
* 2, 4, 6, 8 or for 4 the user is allowed to enter 1, 5, 7. IF the security code to
* enter is 1478 and if the user enters 1178 he should be allowed. Write a function
* to take security code from the user and print out if he should be allowed or not */这题难度并不高，如果在leetcode里面，就是个简单题水平。。而且因为最多检查一次是否是该数字上下左右，剩下的时候只要单纯比较是不是一样就行了。。
那唯一一次比较我的处理是把user输入和原密码在keypad的位置计算出来，然后比较他们是不是相邻关系。。
代码如下：
# include
# include
using namespace std;
const vector pwd = {1, 4, 7, 8};
const vector> keypad = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
bool pwd_allowed(vector);
int main(int argc, char** argv) {
vector user = {1, 4, 7, 8};
cout << "True or False: " << pwd_allowed(user) << endl;
return 0;
}
bool pwd_allowed(vector user) {
bool mistake = false;
for (int i = 0; i < user.size(); ++i) {
if (user[i] == pwd[i])
continue;
else {
if (mistake == true)
return false;
else {
mistake = true;
int p_row;
int p_col;
int u_row;
int u_col;
/* find the posistion of the pwd[i] */
for (int j = 0; j < keypad.size(); ++j) {
for (int k = 0; k < keypad[0].size(); ++k) {
if (pwd[i] == keypad[j][k]) {
p_row = j;
p_col = k;
}
if (user[i] == keypad[j][k]) {
u_row = j;
u_col = k;
}
}
}
/* 4 condition */
if ((p_row == u_row and p_col == u_col - 1)
or (p_row == u_row and p_col == u_col + 1)
or (p_col == u_col and p_row == u_row - 1)
or (p_col == u_col and p_row == u_row + 1))
;
else
return false;
}
}
}
return true;
} Epic 05 Continuous alphabets
题目要求如下：
/* Description: Print continuous alphabets from a sequence of arbitrary alphabets
* For example:
* Input: abcdefljdflsjflmnopflsjflasjftuvwxyz
* Output: abcdef; mnop; tuvwxyz
* Input: AbcDefljdflsjflmnopflsjflasjftuvWxYz
* Output: abcdef; mnop; tuvwxyz */如果和前面的字母相同，则在str 加上这个字母，如果不同，则看是否str已经存储大于1个元素，如果大于1个，则将其存入vector中，否则不作处理，但是无论如何都要以当前的char建立一个新的str
详细代码如下：
# include
# include
# include
using namespace std;
vector con_alpha(string);
int main(int argc, char** argv) {
string input = "AbcDefljdflsjflmnopflsjflasjftuvWxYz";
for (int i = 0; i < input.length(); ++i)
input[i] = tolower(input[i]);
cout << input << endl;
vector res = con_alpha(input);
for (int i = 0; i < res.size(); ++i)
cout << res[i] << endl;
return 0;
}
vector con_alpha(string input) {
vector res;
int i = 1;
string str = string(1, input[0]);
while (i < input.length()) {
char prev = input[i-1];
char cur = input[i];
if (prev + 1 == cur) {
str += cur;
}
else {
if (str.length() > 1) {
res.push_back(str);
}
str = string(1, input[i]);
}
++i;
}
if (str.length() > 1)
res.push_back(str);
return res;
} Epic 06 Palindromes
本题题目要求如下：
/* Print all palindromes of size greater than equal to 3 of a given string. (DP) */这题的dp算法我没有想出来，是参考一个网友的答案，在careercup上id是XiaoPiGU。。。这名字真让人无语。。。
要想让[i,j]是palindrome，则条件是：[i+1,j-1]是palindrom并且[i] == [j]，该dp算法就是根据这个做的
pal[j-1] == 1就意味着[1+1,j-1]是palindrome，再判断[i]和[j],如果想等，则palin[j] = 1，且[i,j]为palindrome，否则palin[j] = 0
代码如下：
int main(int argc, char** argv) {
string str = "ababa";
vector res = Palindrome(str);
for (int i = 0; i < res.size(); ++i)
cout << res[i] << endl;
return 0;
}
vector Palindrome(string str) {
vector res;
vector pal(str.size(), 0);
if (str.size() < 3)
return res;
else {
for (int i = str.size() - 1; i >= 0; --i) {
for (int j = str.size() - 1; j >= i; --j) {
if (str[i] == str[j] and (j - i <= 1 or pal[j-1] == 1)) {
pal[j] = 1;
if (j - i > 1)
res.push_back(str.substr(i, j - i + 1));
}
else
pal[j] = 0;
}
}
}
return res;
} 本题题目要求如下：
/* Description: User inputs a series of numbers and terminates the series by a zero.
* Your program should find the first three maximum values in the series and exclude them
* from the series and compute the average of the remaining numbers. (excluding zero as well)
* Ex - 3, 7, 12, 2, 25, 8, 9, 13, 10, 0
* First three maximum numbers = 25, 13, 12
* Average of the rest = (3 + 7 + 2 + 8 + 9 + 10) / 6 = 6.5 */# include
# include
using namespace std;
double average();
int main(int argc, char** argv) {
cout << average() << endl;
return 0;
}
double average() {
int tmp;
int max1 = INT_MIN;
int max2 = INT_MIN;
int max3 = INT_MIN;
int count = 0;
int sum = 0;
while (cin >> tmp) {
if (tmp == 0)
break;
else {
++count;
sum += tmp;
if (tmp > max1) {
max1 = tmp;
if (tmp > max2) {
swap(max1, max2);
if (tmp > max3) {
swap(max2, max3);
}
}
}
}
}
double res = 1.0 * (sum - max1 - max2 - max3) / (count - 3);
cout << max1 << ", " << max2 << ", " << max3 << endl;
return res;
} Epic 08 Triangle
题目要求如下：
/* Description: Given a array
* {{ 4,7,3,6,7}}
* construct a triangle like
* {{81}}
* {{40,41}}
* {{21,19,22}}
* {{11,10,9,13}}
* {{ 4,7,3,6,7}} */
下面说明了过程：
recur(level 5) {
recur(level 4);
push(level 5);
}
详细代码如下：
# include
# include
using namespace std;
void triangle(vector);
vector> res;
int main(int argc, char** argv) {
vector arr = {4, 7, 3, 6, 7};
triangle(arr);
for (int i = 0; i < res.size(); ++i) {
for (int j = 0; j < res[i].size(); ++j) {
cout << res[i][j] << " ";
}
cout << endl;
}
}
void triangle(vector arr) {
if (arr.size() == 0)
return;
else if (arr.size() > 1) {
vector new_arr(arr.size() - 1);
for (int i = 0; i < new_arr.size(); ++i)
new_arr[i] = arr[i] + arr[i+1];
triangle(new_arr);
}
res.push_back(arr);
} Epic 09 Print stars
本题题目要求如下：
/* Write a software to print triangle made of *s. Given the height and width of Triangles in terms of number of stars.
* like to output
* *
* * *
* * * *
* given you have to use 3 stars or the height is 3 stars. */我有点不敢相信这是面经中的题目。。。。谁碰到这道题真算是中奖了。。。
另外上面的star别混了。每行第一个star都是注释的标志。。
代码如下：
# include
# include
using namespace std;
vector> star(int);
int main(int argc, char** argv) {
int height = 4;
vector> res = star(height);
for (int i = 0; i < res.size(); ++i) {
for (int j = 0; j < res[i].size(); ++j) {
cout << res[i][j];
}
cout << endl;
}
}
vector> star(int height) {
vector> ret;
for (int i = 1; i <= height; ++i) {
vector tmp(i, '*');
ret.push_back(tmp);
}
return ret;
} Epic 10 Seed Number
题目要求如下：
/* Description: Find the seed of a number.
* Eg : 1716 = 143*1*4*3 =1716 so 143 is the seed of 1716. find all possible seed for a given number. */本题我当时乍一看也不会，但其实很简单，就是从sqrt(num) - num之间找一个数字，每个都检测，它本身乘以每位的数字，看看积是不是原数，如果是就返回，如果不是，就继续找，代码如下：
# include
# include
using namespace std;
int seed_number(int);
int main(int argc, char** argv) {
int num = 1716;
int res = seed_number(num);
cout << res << endl;
}
int seed_number(int num) {
for (int seed = sqrt(num); seed < num; ++seed) {
if (num % seed == 0) {
int product = seed;
int tmp = seed;
while (tmp != 0) {
product *= (tmp % 10);
tmp /= 10;
}
if (product == num)
return seed;
}
}
} Epic 11 Permutations of numbers
本题题目要求如下：
/* Description: Length is given as input.Print all possible permutations of numbers between 0-9.
* Eg: if input length=4
* all possible combinations can be 0123, 1234, 5678,9864,...etc all combinations of length from
* in all numbers between 0-9 */个人认为这道题就跟所有的backtracking，或者dfs算法一样。。。
代码如下：
# include
# include
using namespace std;
vector> res;
void permutation(vector, vector, int);
int main(int argc, char** argv) {
vector digits = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
vector tmp;
permutation(tmp, digits, 2);
for (int i = 0; i < res.size(); ++i) {
for (int j = 0; j < res[i].size(); ++j) {
cout << res[i][j];
}
cout << endl;
}
return 0;
}
void permutation(vector str, vector digits, int n) {
if (str.size() == n) {
res.push_back(str);
return;
}
else {
for (int i = 0; i < digits.size(); ++i) {
int tmp = digits[i];
digits.erase(digits.begin() + i);
str.push_back(tmp);
permutation(str, digits, n);
str.pop_back();
digits.insert(digits.begin() + i, tmp);
}
}
} Epic 12 Snake sequence
本题题目要求如下：
/* You are given a grid of numbers. A snake sequence is made up of adjacent numbers
* such that for each number, the number on the right or the number below it is +1 or -1 its value.
* For example,
* 1 3 2 6 8
* -9 7 1 -1 2
* 1 5 0 1 9
* In this grid, (3, 2, 1, 0, 1) is a snake sequence.
* Given a grid, find the longest snake sequences and their lengths
* (so there can be multiple snake sequences with the maximum length). */本题耗费了我不少时间，是用dp做的，dp[i][j]存储以访问过的节点为起点，该点为终点的最长的序列。。所以dp[i][j]就应该是d[i-1][j]加上该位置元素，dp[i][j-1]加上该位置元素这两者的最大值。。。这两者的存在条件分别是[i-1][j]与[i][j]只差1，[i][j-1]与[i][j]只差一，如果这种情况都不存在，则dp[i][j]只包含[i][j]这一个元素形成的序列。。
该动态规划算法代码如下：
# include
# include
# include
# include
using namespace std;
const vector> input = {
{1, 3, 2, 6, 8},
{-9, 7, 1, -1, 2},
{1, 5, 0, 1, 9}
};
vector find_snake();
int main(int argc, char** argv) {
vector res = find_snake();
for (int i = 0; i < res.size(); ++i)
cout << res[i] << " ";
cout << endl << "length: " << res.size() << endl;
return 0;
}
vector find_snake() {
vector longest;
vector>> sequence(input.size(), vector>(input[0].size(), vector()));
for (int i = 0; i < input.size(); ++i) {
for (int j = 0; j < input[0].size(); ++j) {
sequence[i][j].push_back(input[i][j]);
if (i > 0 and abs(input[i][j] - input[i-1][j]) == 1) {
if (sequence[i][j].size() < sequence[i-1][j].size() + 1) {
sequence[i][j].assign(sequence[i-1][j].begin(), sequence[i-1][j].end());
sequence[i][j].push_back(input[i][j]);
}
}
if (j > 0 and abs(input[i][j] - input[i][j-1]) == 1) {
if (sequence[i][j].size() < sequence[i][j-1].size() + 1) {
sequence[i][j].assign(sequence[i][j-1].begin(), sequence[i][j-1].end());
sequence[i][j].push_back(input[i][j]);
}
}
if (sequence[i][j].size() > longest.size())
longest.assign(sequence[i][j].begin(), sequence[i][j].end());
}
}
return longest;
} Epic 13 Unique Number
题目要求如下：
/* Description: Write, efficient code for extracting unique elements from
* a sorted list of array. e.g. (1, 1, 3, 3, 3, 5, 5, 5, 9, 9, 9, 9) -> (1, 3, 5, 9). */其实这题我觉得没啥好说的。。就当熟悉一下stl的链表操作吧。。直接上代码：
# include
# include
# include
using namespace std;
list unique(list);
const vector arr = {1, 1, 3, 3, 3, 5, 5, 5, 9, 9, 9, 9};
int main(int argc, char** argv) {
list arr_list(arr.begin(), arr.end());
list res = unique(arr_list);
for (list::iterator it = res.begin(); it != res.end(); ++it)
cout << *it << " ";
cout << endl;
return 0;
}
list unique(list arr_list) {
list res;
for (list::iterator it = arr_list.begin(); it != arr_list.end(); ++it) {
if (*it == res.back())
continue;
else
res.push_back(*it);
}
return res;
}
Epic 14 Valid Password
本题题目要求如下：
/* Description: In 1-9 keypad one key is not working. If some one enters a password then
* not working key will not be entered. You have given expected password and entered password.
* Check that entered password is valid or not
* Ex: entered 164, expected 18684 (you need to take care as when u enter
* 18684 and 164 only both will be taken as 164 input) */先说下要限制的几个条件吧：
1，限制出现过两个不同的元素：只许有一个broken key,所以如果pwd[j] != enter[i], if(pwd[j] != broken) return false
2，限制元素种类相同，但是两个string仍然不同：比如：1246和12646，所以在检测到第一个6，将其设为broken的时候，即使之后的pwd[j] == enter[i]也要检测enter[i]是否等于broken
3，enter提前pwd结束，此时要检测pwd后面的元素是不是都是broken
4，pwd提前结束，那肯定是错的
代码如下：
# include
# include
using namespace std;
bool valid_pwd(string, string);
int main(int argc, char** argv) {
string enter = "1246";
string pwd = "12646";
cout << "Result: " << valid_pwd(enter, pwd) << endl;
return 0;
}
bool valid_pwd(string enter, string pwd) {
int i;
int j;
char broken = 'a';
for (i = 0, j = 0; i < enter.length() and j < pwd.length(); ++i, ++j) {
if (enter[i] == pwd[j]) {
if (broken != 'a' and enter[i] == broken)
return false;
}
else {
if (broken == 'a')
broken = pwd[j];
else {
if (pwd[j] != broken)
return false;
}
--i;
}
}
while (j < pwd.length() and pwd[j] == broken)
++j;
if (i == enter.length() and j == pwd.length())
return true;
else
return false;
} Epic 15 Count and say
本题题目要求如下：
/* Description: Implement LookAndSay function. For example, first, let user input a number,
* say 1. Then, the function will generate the next 10 numbers which satisfy this condition:
* 1, 11,21,1211,111221,312211...
* explanation: first number 1, second number is one 1, so 11.
* Third number is two 1(previous number), so 21. next number one 2 one 1, so 1211 and so on... */# include
# include
# include
using namespace std;
string count_and_say(string);
int main(int argc, char** argv) {
vector arr;
string str = "1";
arr.push_back(str);
int n = 10;
for (int i = 1; i < n; ++i) {
str = count_and_say(str);
arr.push_back(str);
}
for (int i = 0; i < arr.size(); ++i)
cout << arr[i] << endl;
return 0;
}
string count_and_say(string str) {
char tmp = str[0];
int count = 1;
string ret;
for (int i = 1; i < str.length(); ++i) {
if (str[i] == str[i-1]) {
++count;
}
else {
ret.append(to_string(count));
ret += str[i-1];
count = 1;
}
}
ret.append(to_string(count));
ret += str[str.length()-1];
return ret;
} Epic 16 Substring Addition
本题题目要求如下
/* Description: Substring Addition
* Write a program to add the substring
* eg :say you have a list {1 7 6 3 5 8 9 } and user is entering a sum 16.
* Output should display (2-4) that is {7 6 3} cause 7+6+3=16. */# include
# include
using namespace std;
string check(vector, int);
int main(int argc, char** argv) {
int input[] = {1, 7, 6, 3, 5, 8, 9};
vector arr(input, input + sizeof(input)/sizeof(input[0]));
int sum = 16;
string res = check(arr, sum);
cout << res << endl;
return 0;
}
string check(vector arr, int sum) {
for (int i = 0; i < arr.size(); ++i) {
int total = 0;
for (int j = i; j < arr.size(); ++j) {
total += arr[j];
if (total == sum)
return to_string(i+1).append("-").append(to_string(j+1));
}
}
return "Not found";
} Epic 17 A to one
本题题目要求如下：
* Description: From a given string, replace all instances of 'a' with 'one' and 'A' with 'ONE'.
* Example Input:
* " A boy is playing in a garden"
* Example Output:
* " ONE boy is playing in one garden"
* -- Not that 'A' and 'a' are to be replaced only when they are single characters, not as part of another word. */就说说C++怎么解的吧，虽然解法并不严谨。。
我就是用stringstream把该句子parse成多个单词，再替换，组合成新的句子。。
代码如下：
# include
# include
# include
# include
using namespace std;
string a_to_one(string);
int main(int argc, char** argv) {
string str = " A boy is playing in a garden";
string ret = a_to_one(str);
cout << ret << endl;
return 0;
}
string a_to_one(string str) {
stringstream ss(str);
vector res;
string word;
while (ss >> word) {
if (word == "a")
res.push_back("one");
else if (word == "A")
res.push_back("ONE");
else
res.push_back(word);
}
string ret(res[0]);
for (int i = 1; i < res.size(); ++i)
ret.append(" ").append(res[i]);
return ret;
} stringstream感觉挺重要的而且很强大，搞完epic面试，专门研究下用法
Epic 18 Well-ordered String
题目要求如下：
/* Description: You know a password is well-ordered string. Well-ordered string means that
* the order of the characters is in an alphabetical increasing order. Like “abEm” is
* a well-ordered number. However, “abmE” is not a well-order number. Given an input #
* that tells you also how many digits are in the password, print all possible passwords. */本题的算法并不难，就是普通的dfs就能解决问题，代码相当之短，大家看了后就会明白。。
就是注意一个新知识：如果char A = 'a'，
++A就是 'b'
但是怎么到任意元素呢？
A + i是行不通的，至少在我用的gcc 4.8,2是不支持的。。需要强制转换：static_cast
详细代码如下：
# include
# include
# include
using namespace std;
vector res;
void dfs(int, int, string);
int main(int argc, char** argv) {
int remain = 4;
dfs(remain, 0, "");
for (int i = 0; i < res.size(); ++i)
cout << res[i] << endl;
return 0;
}
void dfs(int remain, int start, string prefix) {
if (remain == 0)
res.push_back(prefix);
else {
for (int i = start; i <= 26 - remain; ++i) {
dfs(remain - 1, i + 1, prefix + static_cast('A' + i));
dfs(remain - 1, i + 1, prefix + static_cast('a' + i));
}
}
} Epic 19 Phone numbers
题目要求如下：
/* Description: Print all valid phone numbers of length n subject to following constraints:
* 1.If a number contains a 4, it should start with 4
* 2.No two consecutive digits can be same
* 3.Three digits (e.g. 7,2,9) will be entirely disallowed, take as input */本题仍然我是用dfs来解决的，把三个限制条件加上，并不难。。。基本就是leetcode里面easy题的水准。代码如下：
# include
# include
# include
using namespace std;
vector res;
const string digits("0134568");
void dfs(string, int);
int main(int argc, char** argv) {
int count = 4;
dfs("", count);
for (int i = 0; i < res.size(); ++i)
cout << res[i] << endl;
return 0;
}
void dfs(string num, int count) {
if (num.length() == count)
res.push_back(num);
else {
for (int i = 0; i < digits.length(); ++i) {
if (num.length() > 0) {
if (digits[i] == num[num.length()-1])
continue;
if (digits[i] == '4' and num[0] != '4')
continue;
}
dfs(num + digits[i], count);
}
}
} Epic 20 SMS Problem
题目要求如下：
/* SMS Problem
* 1 - NULL, 2 - ABC, 3 - DEF, 4 - GHI, 5 - JKL, 6 - MON, 7 - PQRS, 8 - TUV, 9 - WXYZ, * - , # -
* We must convert the numbers to text.
* Eg
* I/P - O/P
* 22 - B
* 23 - AD
* 223 - BD
* 22#2 - BA (# breaks the cycle)
* 3#33 - DE
* 2222 - 2
* 2222#2 - 2A
* 22222 - A (cycle must wrap around)
* 222222 - B */ 本题就是单纯的考细心。。。Epic的题感觉挺无聊的。。没啥算法可言，难度基本就是leetcode里面的easy，但是这种题不给调试的机会又很难一次写对。。
代码如下：
# include
# include
# include
using namespace std;
string code[] = {"", "ABC2", "DEF3", "GHI4", "JKL5", "MON6", "PQRS7", "TUV8", "WXYZ9"};
vector hash_map(code, code + sizeof code / sizeof code[0]);
string sms(string);
int main(int argc, char** argv) {
string str("22213#33");
cout << sms(str) << endl;
}
string sms(string str) {
string res;
for (int i = 0; i < str.length(); ++i) {
if (str[i] == '1')
continue;
else if (str[i] == '*')
res += ' ';
else if (str[i] == '#')
continue;
else {
int count = 1;
while (i + 1 < str.length() and str[i] == str[i+1]) {
++count;
++i;
}
res += hash_map[str[i] - '1'][count % (hash_map[str[i] - '1'].size()) - 1];
}
}
return res;
} Epic 21 Keypad Permutation
本题题目要求如下：
/* Description: key pad question - you how your phone has letters on the number keys.
* for example, number 2 has ABC on it, number 3 has DEF, 4 number has GHI,... ,
* and number 9 has WXYZ. Write a program that will print out all of the possible
* combination of those letters depending on the users input. For example,
* say a user presses 234, the output should be
* ADG, ADH, ADI, AEG, AEH, AEI, AFG, AFH, AFI
* BDG, BDH, BDI, BEG, BEH, BEI, BFG, BFH, BFI
* CDG, CDH, CDI, CEG, CEH, CEI, CFG, CFH, CFI */Leetcode上又几乎一模一样的题，我的博客里面有讲解。。叫letter combinations of phone numbers
这里就不多废话了。。。这种combination的题很多都可以用DFS/Backtracking解决，这题也不例外。。。直接上代码：
# include
# include
# include
# include
using namespace std;
vector res;
unordered_map str_map;
void init();
void dfs(string, string, int);
int main(int argc, char** argv) {
string input = "234";
init();
dfs("", input, 0);
for (int i = 0; i < res.size(); ++i)
cout << res[i] << endl;
}
void init() {
str_map['2'] = "ABC";
str_map['3'] = "DEF";
str_map['4'] = "GHI";
str_map['5'] = "JKL";
str_map['6'] = "MNO";
str_map['7'] = "PQRS";
str_map['8'] = "TUV";
str_map['9'] = "WXYZ";
}
void dfs(string tmp, string input, int i) {
if (tmp.size() == input.size())
res.push_back(tmp);
else {
for (int j = 0; j < str_map[input[i]].size(); ++j) {
dfs(tmp + str_map[input[i]][j], input, i + 1);
}
}
} Epic 22 The stepping number
题目要求如下：
/* Description: A number is called as a stepping number if the adjacent digits are having
* a difference of 1. For eg. 8,343,545 are stepping numbers. While 890, 098 are not.
* The difference between a ‘9’ and ‘0’ should not be considered as 1.
* Given start number(s) and an end number(e) your function should list out all the
* stepping numbers in the range including both the numbers s & e. */# include
# include
# include
using namespace std;
vector res;
void dfs(int, int, int, int);
int main(int argc, char** argv) {
int s = 1;
int e = 1000000;
string s_str = to_string(s);
string e_str = to_string(e);
for (int i = s_str.length(); i < e_str.length(); ++i) {
for (int j = 1; j <= 9; ++j)
dfs(s, e, i, j);
}
for (int i = 0; i < res.size(); ++i)
cout << res[i] << endl;
return 0;
}
void dfs(int s, int e, int depth, int tmp) {
if (to_string(tmp).length() == depth) {
if (tmp < s or tmp > tmp)
return;
else
res.push_back(tmp);
}
else {
if (tmp % 10 != 9)
dfs(s, e, depth, tmp * 10 + tmp % 10 + 1);
if (tmp % 10 != 0)
dfs(s, e, depth, tmp * 10 + tmp % 10 - 1);
}
} Epic 23 Goldbach's conjecture
本题的背景是哥德巴赫猜想。。其实要不是这道题。。我都不知道哥德巴赫猜想是什么。。。
题目要求如下：
/* Description: Every even integer greater than 2 can be expressed as the sum of two primes.
* Write a function which takes a number as input, verify if is an even number
* greater than 2 and also print at least one pair of prime numbers. */代码如下：
# include
# include
using namespace std;
bool isPrime(int);
int main(int argc, char** argv) {
for (int check = 4; check <= 200; check += 2) {
for (int i = 2; i <= check / 2; ++i) {
if (isPrime(i) and isPrime(check - i)) {
cout << check << ": " << i << " and " << check - i << endl;
break;
}
}
}
return 0;
}
bool isPrime(int num) {
if (num == 2)
return true;
else {
for (int i = 2; i <= sqrt(num); ++i) {
if (num % i == 0)
return false;
}
}
return true;
} Epic 24 Finding Words
题目要求如下：
/* Description: Write a program for a word search. If there is an NxN grid with one letter
* in each cell. Let the user enter a word and the letters of the word are said to be found
* in the grid either the letters match vertically, horizontally or diagonally in the grid.
* If the word is found, print the coordinates of the letters as output. */本题我仍然使用dfs，每次确定了一个点之后，就对周围八个点进行判断，看是否继续dfs，代码如下，这次我写的虽然很多，但是后面的是init()函数，其实本来不用写这么多废代码的，核心函数就是dfs：
# include
# include
# include
using namespace std;
vector> grid;
vector> res;
void init();
void dfs(int, int, string, vector);
int main(int argc, char** argv) {
init();
for (int i = 0; i < grid.size(); ++i) {
for (int j = 0; j < grid[0].size(); ++j)
cout << grid[i][j] << " ";
cout << endl;
}
string word = "oje";
vector inter_val;
for (int i = 0; i < grid.size(); ++i) {
for (int j = 0; j < grid[0].size(); ++j) {
if (grid[i][j] == word[0]) {
dfs(i, j, word, inter_val);
}
}
}
for (int i = 0; i < res.size(); ++i) {
for (int j = 0; j < res[0].size(); ++j)
cout << res[i][j] << endl;
cout << endl;
}
return 0;
}
void dfs(int i, int j, string word, vector inter_val) {
string tmp;
tmp.append(to_string(i)).append("----").append(to_string(j));
inter_val.push_back(tmp);
if (inter_val.size() == word.size())
res.push_back(inter_val);
else {
if (i > 0 and word[inter_val.size()] == grid[i-1][j])
dfs(i-1, j, word, inter_val);
if (j > 0 and word[inter_val.size()] == grid[i][j-1])
dfs(i, j-1, word, inter_val);
if (i < grid.size() -1 and word[inter_val.size()] == grid[i+1][j])
dfs(i+1, j, word, inter_val);
if (j < grid[0].size() -1 and word[inter_val.size()] == grid[i][j+1])
dfs(i, j+1, word, inter_val);
if (i > 0 and j > 0 and word[inter_val.size()] == grid[i-1][j-1])
dfs(i-1, j-1, word, inter_val);
if (i < grid.size() -1 and j < grid[0].size() -1 and word[inter_val.size()] == grid[i+1][j+1])
dfs(i+1, j+1, word, inter_val);
if (i > 0 and j < grid[0].size() -1 and word[inter_val.size()] == grid[i-1][j+1])
dfs(i-1, j+1, word, inter_val);
if (i < grid.size() -1 and j > 0 and word[inter_val.size()] == grid[i+1][j-1])
dfs(i+1, j-1, word, inter_val);
}
}
void init() {
vector tmp;
tmp.push_back('a');
tmp.push_back('b');
tmp.push_back('c');
tmp.push_back('d');
grid.push_back(tmp);
tmp.clear();
tmp.push_back('e');
tmp.push_back('f');
tmp.push_back('e');
tmp.push_back('h');
grid.push_back(tmp);
tmp.clear();
tmp.push_back('i');
tmp.push_back('j');
tmp.push_back('j');
tmp.push_back('e');
grid.push_back(tmp);
tmp.clear();
tmp.push_back('m');
tmp.push_back('n');
tmp.push_back('o');
tmp.push_back('j');
grid.push_back(tmp);
} Epic 25 Maximum Subarray
本题题目要求如下：
/* Description: Find the largest sum contiguous sub array.
* The length of the returned sub array must be at least of length 2. */代码如下：
# include
# include
# include
# include
using namespace std;
pair max_subarr(int[], int);
int main(int argc, char** argv) {
int arr[] = {-1, -1, 0, 6, -10 };
int n = sizeof arr / sizeof arr[0];
pair res = max_subarr(arr, n);
cout << res.first << " - " << res.second << endl;
return 0;
}
pair max_subarr(int arr[], int n) {
if (n <= 1)
return pair(-1, -1);
int end = 0;
int local_max = arr[0];
int sum = arr[0];
for (int i = 1; i < n; ++i) {
if (local_max + arr[i] >= sum) {
sum = local_max + arr[i];
end = i;
}
local_max = max(local_max + arr[i], arr[i]);
}
int start = end;
int tmp = arr[start];
do {
tmp += arr[--start];
} while (sum != tmp);
return pair(start, end);
} local_max这个变量用来存储以i为终点的subarray的和的最大值，
而sum表示从i = 0到现在的iteration之间最大的subarray的值..
通过第一次迭代，可以求得这个subarray的和以及终点。。
然后通过第二次迭代求起点。。。
这题比较重要！！！！
Epic 26 Subtraction of two Arrays
题目如下：
/* Description: Suppose you want to do the subtraction of two numbers.
* Each digit of the numbers is divided and put in an array. Like A=[1,2, 3, 4, 5],
* B=[4, 5, 3, 5]. You should output an array C=[7, 8, 1, 0].
* Remember that your machine can’t hand numbers larger than 20. */代码如下：
# include
# include
using namespace std;
vector sub(vector, vector);
int main(int argc, char** argv) {
int A[] = {1, 2, 3, 4, 5};
int B[] = {4, 5, 3, 5};
vector A_tmp(A, A + sizeof A / sizeof A[0]);
vector B_tmp(B, B + sizeof B / sizeof B[0]);
vector C = sub(A_tmp, B_tmp);
for (int i = 0; i < C.size(); ++i)
cout << C[i] << " ";
cout << endl;
return 0;
}
vector sub(vector A_tmp, vector B_tmp) {
int i = 0;
int A_n = 0;
while (i < A_tmp.size()) {
A_n = 10 * A_n + A_tmp[i];
++i;
}
i = 0;
int B_n = 0;
while (i < B_tmp.size()) {
B_n = 10 * B_n + B_tmp[i];
++i;
}
int C_n = A_n - B_n;
vector C;
while (C_n != 0) {
C.insert(C.begin(), C_n % 10);
C_n /= 10;
}
return C;
} Epic 27 Additive Number
题目如下：
/* Description: An additive sequence is 1,2,3,5,8,13 where T(n) = T(n -1) + T(n - 2).
* A number range is given to you. Find the additive sequence number in that range.
* Given the start and an ending integer as userinput, generate all integers with
* the following property. */外层循环：选定一个num1和num2.。num1从0开始到i，num1从i开始到j，
然后进入内层循环：看是不是num1+num2就等于剩下的部分，（循环条件：num1 + num2 <= 剩余部分）
(1)如果等于。则返回true
(2)不等于num1 + num2 = num3，检测rest里面，以rest开始为起点，num.length()为长度，的字符串是不是等于num3，
(2-1)，如果等于：num1 = num2, num2 = num3，rest重新取
(2-2)不等于，跳出
代码如下：
# include
# include
# include
using namespace std;
bool additive(string);
int main(int argc, char** argv) {
int start = 112;
int end = 1123;
vector res;
for (int i = start; i <= end; ++i) {
if (additive(to_string(i)))
res.push_back(i);
}
for (int i = 0; i < res.size(); ++i)
cout << res[i] << endl;
return 0;
}
bool additive(string num) {
for (int i = 1; i < num.length(); ++i) {
for (int j = i + 1; j < num.length(); ++j) {
int num1 = stoi(num.substr(0, i));
int num2 = stoi(num.substr(i, j - i));
int index = j;
int rest = stoi(num.substr(index, num.length() - index));
while (num1 + num2 <= rest) {
if (num1 + num2 == rest)
return true;
int num3 = num1 + num2;
string s_3 = to_string(num3);
int len_3 = s_3.length();
if (s_3 == num.substr(index, s_3.length())) {
num1 = num2;
num2 = num3;
index = index + s_3.length();
rest = stoi(num.substr(index, num.length() - index));
}
else
break;
}
}
}
return false;
} Epic 28 Find Max/Min Number
题目要求如下：
/* Description: Take a series of integers as input till a zero is entered.
* Among these given integers, find the maximum of the odd numbers and the
* minimum of the even integers (not including zero) and print them. */代码如下：
# include
# include
# include
using namespace std;
int main(int argc, char** argv) {
int max_odd = INT_MIN;
int min_even = INT_MAX;
int num;
while (cin >> num) {
if (num == 0)
break;
else if (num % 2 == 1) {
max_odd = max(max_odd, num);
}
else {
min_even = min(min_even, num);
}
}
cout << "MAX: " << max_odd << endl;
cout << "MIN: " << min_even << endl;
return 0;
} Epic 29 Edge Detection
题目如下：
/* Description: Two-dimensional array representation of an image can also be represented
* by a one-dimensional array of W*H size, where W represent row and H represent column
* size and each cell represent pixel value of that image. you are also given a threshold X.
* For edge detection, you have to compute difference of a pixel value with each of
* it's adjacent pixel and find maximum of all differences. And finally compare if
* that maximum difference is greater than threshold X. if so, then that pixel is
* a edge pixel and have to display it. */我的理解是假如一个矩阵是
1 2 3
4 9 5
6 7 8
假定threshold是5，则检测到2时。。。最大差距为7，大于threshold，返回2所在的点
检测到4所在的点。。最大差距为5，大于等于threshold，返回4所在的点
检测到9所在的点。。最大差距为7，大于threshold，返回9所在的点。。。
因为是面经，不是真题。。所以题意感觉有点模糊。。。
代码如下：
# include
# include
# include
using namespace std;
void edge_detect(vector, int, int, int);
int main(int argc, char** argv) {
int arr[] = {1,2,3,4,9,5,6,7,8};
vector input(arr, arr + sizeof arr / sizeof arr[0]);
edge_detect(input, 3, 3, 5);
return 0;
}
void edge_detect(vector input, int W, int H, int T) {
for (int i = 0; i < H; ++i) {
for (int j = 0; j < W; ++j) {
int diff = 0;
int index = W * i + j;
/* left */
if (j > 0)
diff = max(diff, abs(input[index] - input[i * W + j - 1]));
/* right */
if (j < W - 1)
diff = max(diff, abs(input[index] - input[i * W + j + 1]));
/* up */
if (i > 0)
diff = max(diff, abs(input[index] - input[(i - 1) * W + j]));
/* down */
if (i < H - 1)
diff = max(diff, abs(input[index] - input[(i + 1) * W + j]));
if (diff >= T)
cout << "Edge: " << index << endl;
}
}
}