普通栈

先进后出的特点

​​​​​​20. 有效的括号

class Solution {
public:unordered_map<char, char> mp = {{')', '('},{']', '['},{'}', '{'}};bool isValid(string s) {stack<char> st;for (char c : s) {if (c == '(' || c == '[' || c == '{') {st.push(c);} else {if (!st.empty() && mp[c] == st.top()) {st.pop();} else {return false;}}}return st.empty();}
};

155. 最小栈

class MinStack {
private:stack<int> xStack; // 存放正常数据stack<int> minStack; // 顺序存放元素对应的最小值
public:MinStack() {}void push(int val) {xStack.push(val);if (!minStack.empty()) {minStack.push(min(minStack.top(), val));} else {minStack.push(val);}}void pop() {xStack.pop();minStack.pop();}int top() {return xStack.top();}int getMin() {return minStack.top();}
};/*** Your MinStack object will be instantiated and called as such:* MinStack* obj = new MinStack();* obj->push(val);* obj->pop();* int param_3 = obj->top();* int param_4 = obj->getMin();*/

946. 验证栈序列

class Solution {
public:bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {stack<int> st;for (int i = 0, j = 0; i < pushed.size(); ++i) {st.push(pushed[i]);while (!st.empty() && st.top() == popped[j]) {st.pop();++j;}}return st.empty();}
};

394. 字符串解码

class Solution {
public:string getDigits(string& s, int& ptr) {string res;while (isdigit(s[ptr])) {res.push_back(s[ptr++]);}return res;}string getString(vector<string> &v) {string ret;for (const auto &s: v) {ret += s;}return ret;}string decodeString(string s) {vector<string> stk;int ptr = 0;while (ptr < s.size()) {char cur = s[ptr];if (isdigit(cur)) {string digits = getDigits(s, ptr);stk.push_back(digits);} else if (isalpha(cur) || cur == '[') {stk.push_back(string(1, cur));ptr++;} else {ptr++;vector<string> sub;while (stk.back() != "[") {sub.push_back(stk.back());stk.pop_back();}stk.pop_back();int repTIme = stoi(stk.back());stk.pop_back();string subStr;while (!sub.empty()) {subStr += sub.back();sub.pop_back();}string tmp;while (repTIme--) {tmp += subStr;}stk.push_back(tmp);}}return getString(stk);}
};

单调栈

单调栈用途不太广泛，只处理一类典型的问题，比如「下一个更大元素」，「上一个更小元素」等。

739. 每日温度

class Solution {
public:vector<int> dailyTemperatures(vector<int>& temperatures) {// 单调递减栈，即从底到顶逐渐递减stack<int> st;vector<int> res(temperatures.size());for (int i = 0; i< temperatures.size(); ++i) {while (!st.empty() && temperatures[i] > temperatures[st.top()]) {int preIndex = st.top();res[preIndex] = i - preIndex;st.pop();}st.push(i);}return res;}
};

496. 下一个更大元素 I

双数组

class Solution {
public:vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {stack<int> st;unordered_map<int, int> mp;vector<int> res;for (int i = 0; i <  nums2.size(); ++i) {while (!st.empty() && nums2[i] > st.top()) {int preVal = st.top();st.pop();mp[preVal] = nums2[i];}st.push(nums2[i]);}for (auto num : nums1) {if (mp.find(num) != mp.end()) {res.push_back(mp[num]);} else {res.push_back(-1);}}return res;}
};

503. 下一个更大元素 II

循环数组

class Solution {
public:vector<int> nextGreaterElements(vector<int>& nums) {int len = nums.size();stack<int> st;vector<int> res(len, -1);for (int i = 0; i <  2 * len; ++i) {int index = i % len;while (!st.empty() && nums[index] > nums[st.top()]) {int preIndex= st.top();st.pop();res[preIndex] = nums[index];}st.push(index);}return res;}
};

84. 柱状图中最大的矩形

class Solution {
public:int largestRectangleArea(vector<int>& heights) {int n = heights.size();stack<int> st;// 假设 h=heights[i]是矩形的高度，那么矩形的宽度最大是多少？我们需要知道：// 1. 在 i 左侧的小于 h 的最近元素的下标 left// 2. 在 i 右侧的小于 h 的最近元素的下标 rightvector<int> left(n, 0), right(n, n);for (int i = 0; i < n; ++i) {while (!st.empty() && heights[i] < heights[st.top()]) {int index = st.top();st.pop();right[index] = i;}left[i] = st.empty() ? -1 : st.top();st.push(i);}int res = 0;for (int i = 0; i < n; ++i) {int height = heights[i];int wide = right[i] - left[i] - 1;res = max(res, height * wide);}return res;}
};

42. 接雨水

单调栈解法：

【Leetcode每日打卡】接雨水问题的超完全手册 - 知乎 (zhihu.com)

class Solution {
public:int trap(vector<int>& height) {stack<int> st;int res = 0;for (int i = 0; i < height.size(); ++i) {while (!st.empty() && height[i] > height[st.top()]) {int bottomIndex = st.top();st.pop();while (!st.empty() && height[bottomIndex] == height[st.top()]) {st.pop();}if (!st.empty()) {int heightDiff = min(height[st.top()], height[i]) - height[bottomIndex];int wide = i - st.top() - 1;res += heightDiff * wide;}}st.push(i);}return res;}
};

动态规划解法：找到第i个元素左右最大的值得较小值，减去当前高度，即为该列能存储的雨水

class Solution {
public:int trap(vector<int>& height) {int len = height.size();int sum = 0;vector<int> left_h(height.size());vector<int> right_h(height.size());for (int i = 1; i < len; ++i) {left_h[i] = max(left_h[i-1], height[i-1]);}for (int j = len - 2; j >= 0; --j) {right_h[j] = max(right_h[j+1], height[j+1]);}for (int i = 1; i < len - 1; ++i) {int min_h = min(left_h[i], right_h[i]);if (min_h > height[i]) {sum += min_h - height[i];}}return sum;}
};

双指针解法：

class Solution {
public:int trap(vector<int>& height) {int len = height.size();int sum = 0;int left = 0, right = len - 1;int left_max = 0, right_max = 0;while (left < right) {if (height[left] < height[right]) {if (left_max <= height[left]) {left_max = height[left];} else {sum += left_max - height[left];}++left;} else {if (right_max <= height[right]) {right_max = height[right];} else {sum += right_max - height[right];}--right;}}return sum;}
};