描述 Description

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.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

[Min Stack]

分析 Analysis

O(1)时间返回栈中的最小值。

1

使用两个栈来实现，一个栈来按顺序存储push进来的数据，另一个用来存出现过的最小值。

2

只用一个栈，还需要一个整型变量min_val来记录当前最小值，初始化为整型最小值，然后如果需要进栈的数字小于等于当前最小值min_val，那么将min_val压入栈，并且将min_val更新为当前数字。在出栈操作时，先将栈顶元素移出栈，再判断该元素是否和min_val相等，相等的话我们将min_val更新为新栈顶元素，再将新栈顶元素移出栈即可

[[LeetCode] Min Stack 最小栈]

[https://discuss.leetcode.com/topic/7020/java-accepted-solution-using-one-stack

https://discuss.leetcode.com/topic/18556/c-using-two-stacks-quite-short-and-easy-to-understand]

代码 Code

方法1

class MinStack {
public:
    /** initialize your data structure here. */
    MinStack() {}

    void push(int x) {
        s1.push(x);
        if (s2.empty() || x <= s2.top()) s2.push(x);
    }

    void pop() {
        if (s1.top() == s2.top()) s2.pop();
        s1.pop();
    }

    int top() {
        return s1.top();
    }

    int getMin() {
        return s2.top();
    }

private:
    stack<int> s1, s2;
};

方法2

class MinStack {
public:
    /** initialize your data structure here. */
    MinStack() {
        min_val = INT_MAX;
    }

    void push(int x) {
        if (x <= min_val) {
            st.push(min_val);
            min_val = x;
        }
        st.push(x);
    }

    void pop() {
        int t = st.top(); st.pop();
        if (t == min_val) {
            min_val = st.top(); st.pop();
        }
    }

    int top() {
        return st.top();
    }

    int getMin() {
        return min_val;
    }
private:
    int min_val;
    stack<int> st;
};