[LeetCode]96. Min Stack带Min函数的栈

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• getMin() -- Retrieve the minimum element in the stack.

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解法：还是用两个栈s1和s2来模拟。s1按顺序保存数据，而s2保存当前最小值。在进栈时，如果s2为空或者当前进栈值不大于s2栈顶值，则也进s2；出栈时，如果当前出栈值恰好为s2的栈顶值，则s2也要出栈。

class MinStack {
public:
    void push(int x) {
        s1.push(x);
        if (s2.empty() || x <= s2.top()) s2.push(x);
    }
    void pop() {
        if (!s1.empty()) {
            if (!s2.empty() && s1.top() == s2.top()) s2.pop();
            s1.pop();
        }
    }
    int top() {
        if (!s1.empty()) return s1.top();
        return 0;
    }
    int getMin() {
        if (!s2.empty()) return s2.top();
        return 0;
    }
private:
    std::stack<int> s1, s2;
};