题目描述
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.
代码实现
/**
* initialize your data structure here.
*/
var MinStack = function() {
this.s1 = [];
this.s2 = [];
};
/**
* @param {number} x
* @return {void}
*/
MinStack.prototype.push = function(x) {
let s2 = this.s2,
s2Len = s2.length,
s1 = this.s1;
let curMin = s2[s2Len-1];
if(curMin < x)
s2.push(curMin);
else
s2.push(x);
s1.push(x);
};
/**
* @return {void}
*/
MinStack.prototype.pop = function() {
let s1 = this.s1,
s1Len = s1.length,
s2 = this.s2,
s2Len = s2.length;
if(s1Len === 0)
return undefined;
s2.pop();
return s1.pop();
};
/**
* @return {number}
*/
MinStack.prototype.top = function() {
let s1 = this.s1,
s1Len = s1.length;
if(s1.length === 0)
return undefined;
return s1[s1Len-1];
};
/**
* @return {number}
*/
MinStack.prototype.getMin = function() {
let s2 = this.s2,
s2Len = s2.length;
if(s2Len === 0)
return undefined;
return s2[s2Len-1];
};
/**
* Your MinStack object will be instantiated and called as such:
* var obj = Object.create(MinStack).createNew()
* obj.push(x)
* obj.pop()
* var param_3 = obj.top()
* var param_4 = obj.getMin()
*/
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