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// @Title: 最小栈 (Min Stack)
// @Author: 15816537946@163.com
// @Date: 2020-06-25 07:15:36
// @Runtime: 24 ms
// @Memory: 7.5 MB
/*
 * @lc app=leetcode.cn id=155 lang=golang
 *
 * [155] 最小栈
 *
 * https://leetcode-cn.com/problems/min-stack/description/
 *
 * algorithms
 * Easy (49.71%)
 * Likes:    264
 * Dislikes: 0
 * Total Accepted:    39.7K
 * Total Submissions: 79.9K
 * Testcase Example:  '["MinStack","push","push","push","getMin","pop","top","getMin"]\n[[],[-2],[0],[-3],[],[],[],[]]'
 *
 * 设计一个支持 push,pop,top 操作,并能在常数时间内检索到最小元素的栈。
 *
 *
 * push(x) -- 将元素 x 推入栈中。
 * pop() -- 删除栈顶的元素。
 * top() -- 获取栈顶元素。
 * getMin() -- 检索栈中的最小元素。
 *
 *
 * 示例:
 *
 * MinStack minStack = new MinStack();
 * minStack.push(-2);
 * minStack.push(0);
 * minStack.push(-3);
 * minStack.getMin();   --> 返回 -3.
 * minStack.pop();
 * minStack.top();      --> 返回 0.
 * minStack.getMin();   --> 返回 -2.
 *
 *
 */

type MinStack struct {
	stack []item
}

type item struct {
	min, x int
}

func Constructor() MinStack {
	return MinStack{}
}

func (this *MinStack) Push(x int) {
	min := x
	if len(this.stack) > 0 && this.GetMin() < x {
		min = this.GetMin()
	}
	this.stack = append(this.stack, item{min, x})
}

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

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

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

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