2. leetcode(哈系表+字符串)
1. leetcode 242 有效的字母异位词
class Solution {
public:
bool isAnagram(string s, string t) {
int record[26] = {0};
for (int i = 0; i < s.size(); i++) {
record[s[i] - 'a']++;
}
for (int i = 0; i < t.size(); i++) {
record[t[i] - 'a']--;
}
for (int i = 0; i < 26; i++) {
if (record[i] != 0)
return false;
}
return true;
}
};
#endif
#if 0
//leetcode 349 两个数组的交集
//unorder_set 中无序不重复
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
unordered_set<int> result;
unordered_set<int> a_set(nums1.begin(), nums1.end());
for (int i : nums2) {
//find返回i的迭代器，如果找不到该元素，则迭代器指向集合最后一个元素之后的位置
//所以如果i 不在a_set, 返回的就是a_set.end();
if (a_set.find(i) != a_set.end()) {
result.insert(i);
}
}
return vector<int>(result.begin(), result.end());
}
};
//使用map
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int, int> map;
vector<int> res;
for (int i = 0; i < nums1.size(); i++) {
map[nums1[i]] = 1;
}
for (int i = 0; i < nums2.size(); i++) {
if (map[nums2[i]] == 1) {
map[nums2[i]] = 0;
res.push_back(nums2[i]);
}
}
return res;
}
};
2. leetcode 202 快乐数
class Solution {
public:
//取数值的各个位上的单数平方之和
int getSum(int n) {
int sum = 0;
while (n) {
sum += (n % 10) * (n % 10);
n /= 10;
}
return sum;
}
bool isHappy(int n) {
unordered_set<int> res;
while (1) {
int sum = getSum(n);
if (sum == 1) {
return true;
}
// 如果这个sum曾经出现过，说明陷入了无限循环中
if (res.find(sum) != res.end()) {
return false;
} else {
res.insert(sum);
}
n = sum;
}
}
};
3. leetcode 1 两数之和
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> res;
for (int i = 0; i < nums.size(); i++) {
auto iter = res.find(target - nums[i]);
if (iter != res.end()) {
return {iter->second, i};
}
res.insert(pair<int, int>(nums[i], i));
}
return {};
}
};
4. leetcode 383 赎金信
//字符统计
class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
int record[26] = {0};
for (int i = 0; i < magazine.length(); i++) {
record[magazine[i] - 'a']++;
}
for (int i = 0; i < ransomNote.length(); i++) {
record[ransomNote[i] - 'a']--;
if (record[ransomNote[i] - 'a'] < 0) {
return false;
}
}
return true;
}
};
5. leetcode 15 三数之和
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
if (nums.size() < 3) return {};
vector<vector<int>> res; // 保存所有不重复的三元组
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > 0) return res; // 第一个数大于0，后面递增，则可能为0
//去重
if (i > 0 && nums[i] == nums[i - 1] ) {
continue;
}
int left = i + 1, right = nums.size() - 1;
while (right > left) {
if (nums[i] + nums[left] + nums[right] > 0) {
right--;
} else if (nums[i] + nums[left] + nums[right] < 0) {
left++;
} else {
res.push_back(vector<int> {nums[i], nums[left], nums[right]});
// 去重逻辑应该在找到第一个三元组之后
while (right > left && nums[right] == nums[right - 1]) right--;
while (right > left && nums[left] == nums[left + 1]) left--;
//找到答案，双指针收缩
left++;
right--;
}
}
}
return res;
}
};
//
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums)
{
int size = nums.size();
if (size < 3) return {}; // 特判
vector<vector<int> >res; // 保存结果（所有不重复的三元组）
std::sort(nums.begin(), nums.end());// 排序（默认递增）
for (int i = 0; i < size; i++) // 固定第一个数，转化为求两数之和
{
if (nums[i] > 0) return res; // 第一个数大于 0，后面都是递增正数，不可能相加为零了
// 去重：如果此数已经选取过，跳过
if (i > 0 && nums[i] == nums[i-1]) continue;
// 双指针在nums[i]后面的区间中寻找和为0-nums[i]的另外两个数
int left = i + 1;
int right = size - 1;
while (left < right)
{
if (nums[left] + nums[right] > -nums[i])
right--; // 两数之和太大，右指针左移
else if (nums[left] + nums[right] < -nums[i])
left++; // 两数之和太小，左指针右移
else
{
// 找到一个和为零的三元组，添加到结果中，左右指针内缩，继续寻找
res.push_back(vector<int>{nums[i], nums[left], nums[right]});
left++;
right--;
// 去重：第二个数和第三个数也不重复选取
// 例如：[-4,1,1,1,2,3,3,3], i=0, left=1, right=5
while (left < right && nums[left] == nums[left-1]) left++;
while (left < right && nums[right] == nums[right+1]) right--;
}
}
}
return res;
}
};
6. leetcode 18 四数之和
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> res;
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size(); i++) {
if (i > 0 && nums[i] == nums[i - 1]) { //去重
continue;
}
for (int j = i + 1; j < nums.size(); j++) {
if (j > i + 1 && nums[j] == nums[j - 1]) { //去重
continue;
}
int left = j + 1, right = nums.size() - 1;
while (left < right) {
if (nums[i] + nums[j] + nums[left] + nums[right] >target) {
right--;
} else if (nums[i] + nums[j] + nums[left] + nums[right] < target) {
left++;
} else {
res.push_back(vector<int> {nums[i], nums[j], nums[left], nums[right]});
left++;
right--;
while (right > left && nums[right] == nums[right + 1]) right--;
while (right > left && nums[left] == nums[left - 1]) left++;
}
}
}
}
return res;
}
};
字符串
7. leetcode 344 反转字符串
class Solution {
public:
void reverseString(vector<char>& s) {
int n = s.size();
for (int left = 0, right = n - 1; left <= right; ++left, --right) {
swap(s[left],s[right]);
}
}
};
8. leetcode 541 反转字符串II
class Solution {
public:
void reverse(string& s, int start, int end) {
for (int i = start, j = end; i < j; i++, j--) {
swap(s[i], s[j]);
}
}
string reverseStr(string s, int k) {
for (int i = 0; i < s.size(); i += 2 * k) {
//每隔2k个字符的前k个字符进行反转
//剩余字符小于2k但大于等于k个,则反转前k个字符
if (i + k < s.size()) {
reverse(s, i, i + k - 1);
continue;
}
//直到剩余字符小于k个，则反转余下全部
reverse(s, i, s.size() - 1);
}
return s;
}
};
//模拟
class Solution {
public:
string reverseStr(string s, int k) {
int n = s.length();
for (int i = 0; i < n; i += 2 * k) {
reverse(s.begin() + i, s.begin() + min(i + k, n));
}
return s;
}
};
9. 剑指Offer 05 替换空格
class Solution {
public:
string replaceSpace(string s) {
int length = s.size();
for (int i = 0; i <= length - 1; i++) {
if (s[i] == ' ') {
s += " ";
}
}
int i = s.size() - 1;
for (int j = length - 1; j >= 0; j--) {
if (s[j] == ' ') {
s[i--] = '0';
s[i--] = '2';
s[i--] = '%';
} else {
s[i--] = s[j];
}
}
return s;
}
};
//
class Solution {
public:
string replaceSpace(string s) {
string res;
for (auto &c : s) { //遍历原字符串
if (c == ' ') {
res.push_back('%');
res.push_back('2');
res.push_back('0');
} else {
res.push_back(c);
}
}
return res;
}
};
10. leetcode 151 翻转字符串里的单词
class Solution {
public:
//字符串反转
void reverse(string& s, int start, int end) {
for (int i = start, j = end; i < j; i++, j--) {
swap(s[i], s[j]);
}
}
//移除冗余空格 双指针法
void removeExtraSpace(string& s) {
int slow = 0, fast = 0;
while (s.size() > 0 && fast < s.size() && s[fast] == ' ') {
fast++;
}
for (; fast < s.size(); fast++) {
//去除字符串内的多余空格
if (fast > 1 && s[fast] == ' ' && s[fast - 1] == s[fast]) {
continue;
} else {
s[slow++] = s[fast];
}
}
if (slow > 1 && s[slow - 1] == ' ') {
s.resize(slow - 1);
} else {
s.resize(slow);
}
}
string reverseWords(string s) {
removeExtraSpace(s);
reverse(s, 0, s.size() - 1);
for (int i = 0; i < s.size(); i++) {
int j = i;
while (j < s.size() && s[j] != ' ') j++;
reverse(s, i, j - 1);
i = j;
}
return s;
}
};
//自行编写对应的函数
class Solution {
public:
string reverseWords(string s) {
// 反转整个字符串
reverse(s.begin(), s.end());
int n = s.size();
int idx = 0;
for (int start = 0; start < n; ++start) {
if (s[start] != ' ') {
// 填一个空白字符然后将idx移动到下一个单词的开头位置
if (idx != 0) s[idx++] = ' ';
// 循环遍历至单词的末尾
int end = start;
while (end < n && s[end] != ' ') s[idx++] = s[end++];
// 反转整个单词
reverse(s.begin() + idx - (end - start), s.begin() + idx);
// 更新start，去找下一个单词
start = end;
}
}
s.erase(s.begin() + idx, s.end());
return s;
}
};
//双端队列
class Solution {
public:
string reverseWords(string s) {
int left = 0, right = s.size() - 1;
// 去掉字符串开头的空白字符
while (left <= right && s[left] == ' ') ++left;
// 去掉字符串末尾的空白字符
while (left <= right && s[right] == ' ') --right;
deque<string> d;
string word;
while (left <= right) {
char c = s[left];
if (word.size() && c == ' ') {
// 将单词 push 到队列的头部
d.push_front(move(word));
word = "";
}
else if (c != ' ') {
word += c;
}
++left;
}
d.push_front(move(word));
string ans;
while (!d.empty()) {
ans += d.front();
d.pop_front();
if (!d.empty()) ans += ' ';
}
return ans;
}
};
//stringstream + 栈
class Solution {
public:
string reverseWords(string s)
{
stack<string> stk;
string res, temp;
istringstream ss(s); // ss与数输入的字符串s绑定
while (ss >> temp) // 以空格为分隔符将其分割
{
stk.push(temp);
stk.push(" ");
}
if(!stk.empty())
stk.pop(); // 弹出最后压入的那个多余空格
while (! stk.empty())// 单词翻转后的新串
{
res += stk.top();
stk.pop();
}
return res;
}
};
11. 剑指Offer58-II 左旋转字符串
class Solution {
public:
string reverseLeftWords(string s, int n) {
reverse(s.begin(), s.begin() + n);
reverse(s.begin() + n, s.end());
reverse(s.begin(), s.end());
return s;
}
};
class Solution {
public:
int reverse_string(string& s, int start, int end) {
for (int i = start; i <= (start + end) / 2; i++) {
char temp = s[i];
s[i] = s[start+end-i];
s[start+end-i] = temp;
}
return 0;
}
string reverseLeftWords(string s, int n) {
int length = s.length();
reverse_string(s, 0, length-1);
reverse_string(s, 0, length-n-1);
reverse_string(s, length-n, length-1);
return s;
}
};
12. leetcode 28 实现strStr()
//暴力破解
class Solution {
public:
int strStr(string haystack, string needle) {
int n = haystack.size(), m = needle.size();
for (int i = 0; i + m <= n; i++) {
bool flag = true;
for (int j = 0; j < m; j++) {
if (haystack[ i + j] != needle[j]) {
flag = false;
break;
}
}
if (flag) {
return i;
}
}
return -1;
}
};
class Solution {
public:
int strStr(string s, string p) {
int n = s.size(), m = p.size();
for(int i = 0; i <= n - m; i++){
int j = i, k = 0;
while(k < m and s[j] == p[k]){
j++;
k++;
}
if(k == m) return i;
}
return -1;
}
};
//KMP算法
class Solution {
public:
int strStr(string haystack, string needle) {
int n = haystack.size(), m = needle.size();
if(m == 0) return 0;
//设置哨兵
haystack.insert(haystack.begin(),' ');
needle.insert(needle.begin(),' ');
vector<int> next(m + 1);
//预处理next数组
for(int i = 2, j = 0; i <= m; i++){
while(j and needle[i] != needle[j + 1]) j = next[j];
if(needle[i] == needle[j + 1]) j++;
next[i] = j;
}
//匹配过程
for(int i = 1, j = 0; i <= n; i++){
while(j and haystack[i] != needle[j + 1]) j = next[j];
if(haystack[i] == needle[j + 1]) j++;
if(j == m) return i - m;
}
return -1;
}
};