LeetCode(155)题解--Min Stack

https://leetcode.com/problems/min-stack/

题目：

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.

思路：

关键在于找最小值要O(1)复杂度，所以空间换时间。一个额外的vector存储递减序入栈的数字。

AC代码：

class MinStack {
public:
    MinStack(){ st_min.push_back(INT_MAX); };
    void push(int x) {
        if (x <= st_min[st_min.size()-1])
            st_min.push_back(x);
        st.push_back(x);
        return;
    }

    void pop() {
        if (st[st.size() - 1] == st_min[st_min.size() - 1]){
            st_min.pop_back();
        }
        st.pop_back();
        return;
    }

    int top() {
        return st[st.size() - 1];
    }

    int getMin() {
        return st_min[st_min.size() - 1];
    }

private:
    vector<int> st;
    vector<int> st_min;
};