2020-07-25
Z 字形变换
AC代码
class Solution {
public:
string convert(string s, int numRows) {
if (numRows <= 1) {
return s;
}
int n = s.size();
string temp[numRows];
int t_numRows = 0;
int p = 0;
while(p < n) {
while(p < n && t_numRows < numRows) {
temp[t_numRows] += s[p];
p++;
t_numRows++;
}
t_numRows = numRows -2;
while (p < n && t_numRows > 0) {
temp[t_numRows] += s[p];
p++;
t_numRows--;
}
}
string res;
for(int i = 0 ; i < numRows; i++) {
res = res + temp[i];
}
return res;
}
};
优化思路
两层while循环多次判断p
-
实际上需要的string数组长度是min(n, numRows)
优化代码
class Solution { public: string convert(string s, int numRows) { if (numRows <= 1) { return s; } int n = int(s.size()); int len = min(numRows, n); vectortemp(len); int t_numRows = 0; bool goingDown = false; for(int i = 0; i < n; i++) { temp[t_numRows] += s[i]; if (t_numRows == 0 || t_numRows == numRows-1) { goingDown = !goingDown; } t_numRows += goingDown ? 1 :-1; } string res; for (int i = 0; i < len; i++) res += temp[i]; return res; } }; 再次优化
可以直接找新旧数列的数字关系，直接计算
优化代码
class Solution { public: string convert(string s, int numRows) { if (numRows <= 1) { return s; } int len_s = int(s.size()); int unit =(2*numRows-2); int n = len_s/unit; int remain = len_s%unit; string res(len_s, 0); for (int i = 0; i < len_s; i++) { int p = 0; if (i%unit == 0) { p = i/unit+1; } else { int r = i%unit + 1,c = i/unit+1; if (r > numRows) { r = unit-r+2; p = 1; } else if (r == numRows) { p = 1-c; } p += n + (n*2)*(r-2) + 2*(c-1) + min(r-1, remain)+1; if (remain > numRows) { p += max(r-(unit-remain+2),0); } } res[p-1] = s[i]; } return res; } };最终成绩
执行用时：8 ms, 在所有 C++ 提交中击败了98.89%的用户
内存消耗：7.7 MB, 在所有 C++ 提交中击败了100.00%的用户
## 75. 颜色分类
AC代码 计数
class Solution {
public:
void sortColors(vector& nums) {
int n[3] = {0};
for(int i : nums) {
n[i]++;
}
int x = 0;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < n[i]; j++) {
nums[j+x] = i;
}
x += n[i];
}
}
};
优化 三指针法
class Solution {
public:
void sortColors(vector& nums) {
int f,t = int(nums.size())-1,m;
f = m = 0;
while (m <= t) {
if (nums[m] == 0) {
swap(nums[m++], nums[f++]);
} else if (nums[m] == 2) {
swap(nums[m], nums[t--]);
} else {
m++;
}
}
}
void xchg(int& a, int& b) {
a = a+b;
b = a-b;
a = a-b;
}
};
129. 求根到叶子节点数字之和
AC代码
class Solution {
public:
int sum = 0;
void go(TreeNode* root, int num) {
if (root->left == NULL && root->right == NULL) {
sum += num*10+root->val;
return;
}
if (root->left != NULL) {
go(root->left, num*10+root->val);
}
if (root->right != NULL) {
go(root->right, num*10+root->val);
}
}
int sumNumbers(TreeNode* root) {
if (root == NULL) {
return 0;
}
go(root, 0);
return sum;
}
};
29. 两数相除
AC代码
class Solution {
public:
unsigned int i2ui(int n) {
return (n<0&&n != -2147483648)?-n:((n == -2147483648) ? 2147483648 : n);
}
int divide(int dividend, int divisor) {
bool neg = (dividend<0)^(divisor<0);
unsigned int a = i2ui(dividend), b = i2ui(divisor);
unsigned int res = 0;
unsigned int tb = b;
unsigned int add = 1;
while((tb & 0x80000000)==0) {
tb <<= 1;
add <<= 1;
}
while (a >= b) {
if (a >= tb) {
res += add;
a -= tb;
}
add >>=1;
tb >>= 1;
}
res = (res > 2147483647 && !neg) ? INT_MAX : res;
int ires = neg ? ((res>2147483648)?INT_MAX:-res) : res;
return ires;
}
};
思路
利用最基本的列竖式法，先转成正数，再计算
优化
不满足题目的
假设我们的环境只能存储 32 位有符号整数的条件类似上面的算法，把所有数转化为负数，再对divisor=0x80000000时特判
优化代码
class Solution {
public:
int nabs(int n) {
return (n > 0)? -n : n;
}
int divide(int dividend, int divisor) {
int neg = ((dividend<0)^(divisor<0));
dividend = nabs(dividend);
divisor = nabs(divisor);
int sub = 1;
if (divisor==INT_MIN) {
return (dividend == INT_MIN) ? 1 : 0;
}
int t_divisor = -divisor;
while((t_divisor & 0x40000000)==0) {
t_divisor <<= 1;
sub <<= 1;
}
int res = 0;
// cout << t_divisor << " " << sub << endl;
while (dividend <= divisor && sub != 0) {
if (dividend <= -t_divisor) {
dividend += t_divisor;
res -= sub;
}
sub >>= 1;
t_divisor >>= 1;
}
if (dividend <= divisor) {
res = (res == INT_MIN)? res : res-1;
// cout << res << endl;
}
res = !neg ? ((res==-2147483648)?INT_MAX:-res) : res;
return res;
}
};
最终成绩
执行用时：0 ms, 在所有 C++ 提交中击败了100.00%的用户
内存消耗：6 MB, 在所有 C++ 提交中击败了100.00%的用户
36. 有效的数独
AC代码
class Solution {
public:
bool isValidSudoku(vector>& board) {
for (int i = 0; i < 9; i++) {
int r[9] = {0};
int c[9] = {0};
int s[9] = {0};
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.') {
r[board[i][j]-'1']++;
}
if (board[j][i] != '.') {
c[board[j][i]-'1']++;
}
}
int a = i/3;
int b = i%3;
for (int ii = 3*a; ii < 3*(a+1); ii++) {
for (int ij = 3*b; ij < 3*(b+1); ij++) {
if (board[ii][ij] != '.') {
s[board[ii][ij]-'1']++;
}
}
}
for (int j = 0; j < 9; j++) {
if (r[j] > 1 || c[j] > 1 || s[j] > 1) {
return false;
}
}
}
return true;
}
};
5. 最长回文子串
AC代码
class Solution {
public:
map> m;
int rb=0,re=0;
string longestPalindrome(string s) {
int n = int(s.size());
if (n <= 0) {
return "";
}
go(s, 0, n);
for (int off = 1; off < n; off++) {
go(s, off, n);
go(s, 0, n-off);
}
while (!m.empty()) {
int sub = m.begin()->first;
int sum = m.begin()->second;
int beg = (sum-sub)/2;
int end = (sum+sub)/2;
if(go(s, beg,end) && ((re-rb) > (end-beg))) break;
}
return s.substr(rb, re-rb);
}
bool go(string& s,int beg, int end) {
int pos = isPalindrome(s, beg, end);
if (pos != beg) {
end -= pos-beg;
beg = pos;
m[end-beg]=end+beg;
return false;
}else {
m.erase(end-beg);
if ((end-beg) > (re-rb)) {
rb = beg;
re = end;
}
return true;
}
}
int isPalindrome(string& s, int beg, int end) {
int res = -1;
for(int i = 0; i < (end-beg)/2; i++) {
if(s[beg+i] != s[end-1 - i] && i > res) res = i;
}
return beg+res+1;
}
};
优化
参考优秀的题解，大致思想是把每个字符作为中心，向左右展开
class Solution {
public:
int l=0,h=0;
string longestPalindrome(string s) {
int n = int(s.size());
if (n <= 1) {
return s;
}
for (int i = 0; i < n; i++) {
i = findLongest(s, i, n);
}
return s.substr(l, h-l+1);
}
int findLongest(const string& s,int i, int n) {
int high = i;
while (high < n-1 && s[high+1] == s[i]) {
high++;
}// 中部字符全部相同
int ans = high;
while (i > 0 && high < n-1 && s[i-1]==s[high+1]) {
i--;
high++;//向两边展开
}
if ((high - i) > h-l) {
h = high;
l = i; //更新最长串的位置
}
return ans;
}
};
62. 不同路径
思路
大佬们都是用dp，而我是推公式，就是这么简单
AC代码
class Solution {
public:
int uniquePaths(int m, int n) {
if (m > n) {
m = m+n;
n = m-n;
m = m-n;
}
int res = n;
if (m < 2) {
return 1;
}
if (m == 2) {
return n;
}
vector v(m-2, 0);
for (int i = 1; i <= n-1; i++) {
v[0] += i;
for (int j = 1; j < m - 2; j++) {
v[j] += v[j-1];
}
}
for (int i = 0; i < m -2; i++) {
res += v[i];
}
return res;
}
};
最终成绩
执行用时：0 ms, 在所有 C++ 提交中击败了100.00%的用户
内存消耗：5.9MB, 在所有 C++ 提交中击败了100.00%的用户
63. 不同路径 II
AC代码
class Solution {
public:
int uniquePathsWithObstacles(vector>& obstacleGrid) {
int n = int(obstacleGrid.size());
int m = int(obstacleGrid[0].size());
bool swap = false;
if (m > n) {
m = m+n;
n = m-n;
m = m-n;
swap = true;
}
vector v(m+1, 0);
long long t_v0 = 1;
for (int i = 0; i < n; i++) {
if ((!swap && obstacleGrid[n-i-1][m-1]) || (swap && obstacleGrid[m-1][n-1-i])) {
v[0] = 0;
} else {
v[0] = t_v0;
}
t_v0 = v[0];
for(int j = 1; j < m; j++) {
if ((!swap && obstacleGrid[n-i-1][m-1-j]) || (swap && obstacleGrid[m-1-j][n-1-i])) {
v[j] = 0;
} else {
v[j] += v[j-1];
}
}
}
return (int)v[m-1];
}
};
优化1
不需要转置，这个问题来自于试错过程中的错误判断
看了题解以后发现自己的代码和它惊人的相似，原来我无师自通学会动规了？？哈哈哈哈
优化1代码
class Solution {
public:
int uniquePathsWithObstacles(vector>& obstacleGrid) {
int n = int(obstacleGrid.size());
int m = int(obstacleGrid[0].size());
vector v(m+1, 0);
long long t_v0 = 1;
for (int i = 0; i < n; i++) {
if (obstacleGrid[n-i-1][m-1]) {
v[0] = 0;
} else {
v[0] = t_v0;
}
t_v0 = v[0];
for(int j = 1; j < m; j++) {
if (obstacleGrid[n-i-1][m-1-j]) {
v[j] = 0;
} else {
v[j] += v[j-1];
}
}
}
return (int)v[m-1];
}
};