剑指 Offer 30. 包含min函数的栈

定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。

示例:

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

提示：

各函数的调用总次数不超过 20000 次

注意：本题与主站 155 题相同：https://leetcode-cn.com/problems/min-stack/

来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/bao-han-minhan-shu-de-zhan-lcof 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

Solution 1

双栈，一个放置数据，一个放置最小值

import java.util.Stack;

class MinStack {

    private Stack<Integer> dataStack;
    private Stack<Integer> minStack;

    /** initialize your data structure here. */
    public MinStack() {
        dataStack = new Stack<>();
        minStack = new Stack<>();
    }

    public void push(int x) {
        dataStack.push(x);
        if (minStack.isEmpty())
            minStack.push(x);
        else
            minStack.push(Integer.min(x, minStack.peek()));
    }

    public void pop() {
        minStack.pop();
        dataStack.pop();
    }

    public int top() {
        return dataStack.peek();
    }

    public int min() {
        return minStack.peek();
    }
}

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack obj = new MinStack();
 * obj.push(x);
 * obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.min();
 */

Solution 2

cpp

#include <stack>

class MinStack
{
public:
    /** initialize your data structure here. */
    std::stack<int> data;
    std::stack<int> min_val;
    MinStack()
    {
    }

    void push(int x)
    {
        data.push(x);
        if (min_val.empty())
        {
            min_val.push(x);
        }
        else
        {
            min_val.push(std::min(min_val.top(), x));
        }
    }

    void pop()
    {

        if (data.empty() || min_val.empty())
            return;
        data.pop();
        min_val.pop();
    }

    int top()
    {
        if (data.empty())
            return -1;
        return data.top();
    }

    int min()
    {
        if (min_val.empty())
            return -1;
        return min_val.top();
    }
};