155. 最小栈

原题

https://leetcode.cn/problems/min-stack/description/

思路

两个栈

复杂度

时间：O(n)
空间：O(n)

Python代码

class MinStack:

    def __init__(self):
        # 当前栈
        self.stack1 = []
        # 最小元素的栈
        self.stack2 = []        

    def push(self, val: int) -> None:
        self.stack1.append(val)
        if len(self.stack2) == 0 or val < self.stack2[-1]:
            self.stack2.append(val)
        else:
            self.stack2.append(self.stack2[-1])
        

    def pop(self) -> None:
        self.stack1.pop()
        self.stack2.pop()
        

    def top(self) -> int:
        return self.stack1[-1]

    def getMin(self) -> int:
        return self.stack2[-1]


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

Go代码

type MinStack struct {
	// 当前栈
	stack1 []int
	// 最小元素的栈
	stack2 []int
}

func Constructor() MinStack {
	return MinStack{}
}

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

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

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

func (this *MinStack) GetMin() int {
	return this.stack2[len(this.stack2)-1]
}

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