本问题对应的 leetcode 原文链接：剑指 Offer 30. 包含min函数的栈

问题描述

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.

限制：

• 各函数的调用总次数不超过 20000 次

解题思路

视频讲解直达： 本题视频讲解

代码实现

时间复杂度：O(1)
额外空间复杂度：O(n)

class MinStack {
    Stack<Integer> stack1;
    Stack<Integer> stack2;

    /** initialize your data structure here. */
    public MinStack() {
        this.stack1 = new Stack();
        this.stack2 = new Stack();
    }

    public void push(int x) {
        stack1.push(x);
        if(stack2.isEmpty() || x <= stack2.peek().intValue()){
            stack2.push(x);
        }
    }

    public void pop() {
        if(!stack1.isEmpty()){
            //Integer, 数值 > 127 I
            if(stack1.peek().intValue() == stack2.peek().intValue()){
                stack2.pop();
            }
            stack1.pop();
        }
    }

    public int top() {
        return stack1.peek();

    }

    public int min() {
        return stack2.peek();
    }
}

Python

class MinStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack1 = []
        self.stack2 = []

    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        self.stack1.append(x)
        if not self.stack2 or x <= self.stack2[-1]:
            self.stack2.append(x)

    def pop(self):
        """
        :rtype: None
        """
        if self.stack1:
            if self.stack1[-1] == self.stack2[-1]:
                self.stack2.pop()
            self.stack1.pop()

    def top(self):
        """
        :rtype: int
        """
        if self.stack1:
            return self.stack1[-1]

    def min(self):
        """
        :rtype: int
        """
        if self.stack2:
            return self.stack2[-1]


# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.min()

C++

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

    }

    void push(int x) {
        stack1.push(x);
        if(stack2.empty() || x <= stack2.top()){
            stack2.push(x);
        }
    }

    void pop() {
        if(!stack1.empty()){
            if(stack1.top() == stack2.top()){
                stack2.pop();
            }
            stack1.pop();
        }
    }

    int top() {
        if(!stack1.empty()){
            return stack1.top();
        }
        return 0;
    }

    int min() {
        if(!stack2.empty()){
            return stack2.top();
        }
        return 0;
    }
private:
    stack<int> stack1, stack2;
};

/**
 * 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();
 */

Go

type MinStack struct {
    stack1 []int
    stack2 []int
}


/** initialize your data structure here. */
func Constructor() MinStack {
    return MinStack{[]int{}, []int{}}
}


func (this *MinStack) Push(x int)  {
    this.stack1 = append(this.stack1, x)
    if len(this.stack2) == 0 || x <= this.stack2[len(this.stack2)-1] {
        this.stack2 = append(this.stack2, x)
    }
}


func (this *MinStack) Pop()  {
    if len(this.stack1) > 0 {
        if this.stack1[len(this.stack1)-1] == this.stack2[len(this.stack2)-1] {
            this.stack2 = this.stack2[:len(this.stack2)-1]
        }
        this.stack1 = this.stack1[:len(this.stack1)-1]
    }
}


func (this *MinStack) Top() int {
    if len(this.stack1) > 0 {
        return this.stack1[len(this.stack1)-1]
    }
    return 0
}


func (this *MinStack) Min() int {
    if len(this.stack2) > 0 {
        return this.stack2[len(this.stack2)-1]
    }
    return 0
}


/**
 * Your MinStack object will be instantiated and called as such:
 * obj := Constructor();
 * obj.Push(x);
 * obj.Pop();
 * param_3 := obj.Top();
 * param_4 := obj.Min();
 */

JS

/**
 * initialize your data structure here.
 */
var MinStack = function() {
    this.stack1 = [];
    this.stack2 = [];
};

/** 
 * @param {number} x
 * @return {void}
 */
MinStack.prototype.push = function(x) {
    this.stack1.push(x);
    if (this.stack2.length === 0 || x <= this.stack2[this.stack2.length-1]) {
        this.stack2.push(x);
    }
};

/**
 * @return {void}
 */
MinStack.prototype.pop = function() {
    if (this.stack1.length > 0) {
        if (this.stack1[this.stack1.length-1] === this.stack2[this.stack2.length-1]) {
            this.stack2.pop();
        }
        this.stack1.pop();
    }
};

/**
 * @return {number}
 */
MinStack.prototype.top = function() {
    return this.stack1[this.stack1.length-1];
};

/**
 * @return {number}
 */
MinStack.prototype.min = function() {
    return this.stack2[this.stack2.length-1];
};

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

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