LeetCode Hot 100：栈

20. 有效的括号

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

        return stk.empty();
    }
};

155. 最小栈

class MinStack {
private:
    stack<int> s, minStack;

public:
    MinStack() {}

    void push(int val) {
        s.push(val);
        if (minStack.empty() || val <= minStack.top())
            minStack.push(val);
    }

    void pop() {
        int x = s.top();
        s.pop();
        if (!minStack.empty() && minStack.top() == x)
            minStack.pop();
    }

    int top() { return s.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();
 */

394. 字符串解码

class Solution {
public:
    string decodeString(string s) {
        int n = s.size();
        int i = 0;
        stack<string> stk;

        while (i < n) {
            char c = s[i];
            if (isdigit(c)) {
                string num;
                while (isdigit(s[i])) {
                    num.push_back(s[i]);
                    i++;
                }
                stk.push(num);
            } else if (isalpha(c) || c == '[') {
                stk.push(string(1, s[i]));
                i++;
            } else { // c == ']'
                vector<string> sub;
                while (stk.top() != "[") {
                    sub.insert(sub.begin(), stk.top());
                    stk.pop();
                }
                stk.pop(); // 左括号出栈
                // 此时栈顶为当前 sub 对应的字符串应该出现的次数
                int repeatTime = stoi(stk.top());
                stk.pop();
                string temp;
                while (repeatTime--) {
                    for (string& str : sub)
                        temp += str;
                }
                stk.push(temp);
                i++;
            }
        }

        string ans;
        while (!stk.empty()) {
            ans.insert(0, stk.top());
            stk.pop();
        }

        return ans;
    }
};

739. 每日温度

思路 1：单调栈

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int n = temperatures.size();
        vector<int> ans(n, 0);
        stack<int> s;

        for (int i = 0; i < n; i++) {
            int cur = temperatures[i];
            while (!s.empty() && cur > temperatures[s.top()]) {
                int j = s.top();
                s.pop();
                ans[j] = i - j;
            }
            s.push(i);
        }

        return ans;
    }
};

84. 柱状图中最大的矩形

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int n = heights.size();
        // left[i] = 在 i 左侧的小于 heights[i] 的最近元素的下标
        vector<int> left(n, -1);
        stack<int> s;
        for (int i = 0; i < n; i++) {
            while (!s.empty() && heights[i] <= heights[s.top()])
                s.pop();
            if (!s.empty())
                left[i] = s.top();
            s.push(i);
        }
        // left[i] = 在 i 右侧的小于 heights[i] 的最近元素的下标
        vector<int> right(n, n);
        s = stack<int>();
        for (int i = n - 1; i >= 0; i--) {
            while (!s.empty() && heights[i] <= heights[s.top()])
                s.pop();
            if (!s.empty())
                right[i] = s.top();
            s.push(i);
        }

        int ans = INT_MIN;
        for (int i = 0; i < n; i++) {
            int len = right[i] - left[i] - 1;
            ans = max(ans, heights[i] * len);
        }

        return ans;
    }
};