之前总结过字节跳动TOP50算法面试题：

字节跳动常见算法面试题top50整理_沉迷单车的追风少年-CSDN博客_字节算法面试题

目录

双指针

42.接雨水 

283.移动零 

11.盛最多水的容器

15.三数之和 

哈希

1. 两数之和

49.字母异位词分组

128.最长连续序列 

滑动窗口

3.无重复字符的最长子串

 438.找到字符串中所有字母异位词

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双指针

42.接雨水 

用单调栈再做一遍吧……

class Solution {
public:int trap(vector<int>& height) {//当前柱子如果小于等于栈顶元素，说明形不成凹槽，则将当前柱子入栈；//反之若当前柱子大于栈顶元素，说明形成了凹槽，于是将栈中小于当前柱子的元素pop出来，将凹槽的大小累加到结果中。int ans = 0;stack<int> sta;for (int i = 0; i < height.size(); i++) {while (!sta.empty() && height[i] > height[sta.top()]) {int top = sta.top();sta.pop();if (sta.empty()) {break;}int left = sta.top();int w = i - left - 1;int h = min(height[i], height[left]) - height[top];ans += w * h;}sta.push(i);}return ans;}
};

283.移动零 

class Solution {
public:void moveZeroes(vector<int>& nums) {// 双指针// 左指针指向当前已经处理好的序列的尾部// 右指针指向待处理序列的头部// 每次交换都是将左指针的零与右指针的非零数交换，且非零数的相对顺序并未改变int left = 0;int right = 0;while (right < nums.size()) {if (nums[right] != 0) {swap(nums[left], nums[right]);left ++;}right ++;}return;}
};

11.盛最多水的容器

class Solution {
public:int maxArea(vector<int>& height) {int left = 0;int right = height.size() - 1;int ans = 0;while (left < right) {int minheight = min(height[left], height[right]); //盛水取决于最短的线ans = max(ans, (minheight * (right - left)));if (height[left] < height[right]) { // 移动短的线left ++;} else {right --;}}return ans;}
};

15.三数之和 

class Solution {
public:vector<vector<int>> threeSum(vector<int>& nums) {// 排序后用双指针查找vector<vector<int> > ans;sort(nums.begin(), nums.end());int left, right;for (int i = 0; i < nums.size() - 1; i++) {if (nums[i] > 0) {continue;}left = i + 1;right = nums.size() - 1;while (left < right) {if ((nums[i] + nums[left] + nums[right]) == 0) {ans.push_back({nums[i], nums[left], nums[right]});// 去除重复元素while (left < right && nums[left] == nums[left + 1]) {left ++;}while (left < right && nums[right] == nums[right - 1]) {right --;}// 移动指针left ++;right --;} else if ((nums[i] + nums[left] + nums[right]) > 0) {right --;} else {left ++;}}// i 下标也需要去重while (i + 1 < nums.size() && nums[i] == nums[i + 1])i++;}        return ans;}
};

哈希

1. 两数之和

class Solution {
public:vector<int> twoSum(vector<int>& nums, int target) {vector<int> ans;unordered_map<int, int> m;for (int i = 0; i < nums.size(); i++) {if (m.find(nums[i]) != m.end()) {return {m[nums[i]], i};}m[target - nums[i]] = i;}return ans;}
};

49.字母异位词分组

class Solution {
public:vector<vector<string>> groupAnagrams(vector<string>& strs) {vector<vector<string>> ans;// 哈希表存储unordered_map<string, vector<string> > m; for (auto s : strs) {string key = s;// 排序后所有异位词存在唯一keysort(key.begin(), key.end());m[key].push_back(s);}for (auto it = m.begin(); it != m.end(); it++) {ans.push_back(it->second);}return ans;}
};

128.最长连续序列 

class Solution {
public:int longestConsecutive(vector<int>& nums) {// 用哈希表存储是否出现过的数字unordered_set<int> hash;int ans = 0;for (auto num : nums) {hash.insert(num);}for (auto x : hash) {// 如果当前数字前面没有出现过数字,则从当前位置开始查// 因为如果前一个数字出现过, 那么他已经被查找过了if (!hash.count(x - 1)) {int y = x;// 一直往后查找连续的序列while (hash.count(y + 1)) {y++;}// 维护最大的ansans = max(ans, y - x + 1);}}return ans;}
};

滑动窗口

3.无重复字符的最长子串

class Solution {
public:int lengthOfLongestSubstring(string s) {// 滑动窗口int left = 0;int right = 0;int ans = 0;// 记录窗口unordered_map<char, int> windows;while (right < s.size()) {char c = s[right];right ++;windows[c] ++;// 收缩窗口while (windows[c] > 1) {char d = s[left];left ++;windows[d] --;}// 维护最大ansans = max(right - left, ans);}return ans;}
};

 438.找到字符串中所有字母异位词

class Solution {
public:vector<int> findAnagrams(string s, string p) {vector<int> ans;// 滑动窗口int left = 0;int right = 0;int count = 0;// 当前窗口中的字符和需要凑齐的字符unordered_map<char, int> windows, need;// 需要凑齐的字符for (auto c : p) {need[c] ++;}while (right < s.size()) {// 向右滑动char c = s[right];right ++;// 进行窗口内数据更新if (need.count(c)) {windows[c] ++;if (windows[c] == need[c]) {count ++;}}// 左侧收缩窗口while (right - left >= p.size()) {if (count == need.size()) {ans.push_back(left);}char d = s[left];left ++;// 窗口内数据更新if (need.count(d)) {if (windows[d] == need[d]) {count --;}windows[d] --;}}}return ans;}
};