155. 最小栈
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155. 最小栈
Description
Difficulty: 中等
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。
实现 MinStack
类:
MinStack()
初始化堆栈对象。void push(int val)
将元素val推入堆栈。void pop()
删除堆栈顶部的元素。int top()
获取堆栈顶部的元素。int getMin()
获取堆栈中的最小元素。
示例 1:
输入:
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
输出:
[null,null,null,null,-3,null,0,-2]
解释:
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.
提示:
- -231 <= val <= 231 - 1
pop
、top
和getMin
操作总是在 非空栈 上调用push
,pop
,top
, andgetMin
最多被调用 3 * 104 次
Solution
Language: JavaScript
// stack正常push,min_stack只会push需要入栈和栈顶中较小的元素
// stack正常pop,min_stack正常pop
// 返回stack栈顶元素
// 返回min_stack栈顶元素
var MinStack = function() {
this.stack = [];
this.min_stack = [Infinity]
};
MinStack.prototype.push = function(x) {
this.stack.push(x)
this.min_stack.push(Math.min(x, this.min_stack[this.min_stack.length - 1]))
};
MinStack.prototype.pop = function() {
this.stack.pop()
this.min_stack.pop()
};
MinStack.prototype.top = function() {
return this.stack[this.stack.length - 1]
};
MinStack.prototype.getMin = function() {
return this.min_stack[this.min_stack.length - 1]
};