面试题 03.01. 三合一

三合一。描述如何只用一个数组来实现三个栈。

你应该实现push(stackNum, value)、pop(stackNum)、isEmpty(stackNum)、peek(stackNum)方法。stackNum表示栈下标，value表示压入的值。

构造函数会传入一个stackSize参数，代表每个栈的大小。

示例1:

 输入：
["TripleInOne", "push", "push", "pop", "pop", "pop", "isEmpty"]
[[1], [0, 1], [0, 2], [0], [0], [0], [0]]
 输出：
[null, null, null, 1, -1, -1, true]
说明：当栈为空时`pop, peek`返回-1，当栈满时`push`不压入元素。

示例2:

 输入：
["TripleInOne", "push", "push", "push", "pop", "pop", "pop", "peek"]
[[2], [0, 1], [0, 2], [0, 3], [0], [0], [0], [0]]
 输出：
[null, null, null, null, 2, 1, -1, -1]

题解:

/**
 * @param {number} stackSize
 */
var TripleInOne = function (stackSize) {
    this.size = stackSize
    this.stack = []
};

/** 
 * @param {number} stackNum 
 * @param {number} value
 * @return {void}
 */
TripleInOne.prototype.push = function (stackNum, value) {
    if (!this.stack[stackNum]) {
        this.stack[stackNum] = []
    }
    if (this.stack[stackNum].length < this.size) {
        this.stack[stackNum].push(value)
    }
};

/** 
 * @param {number} stackNum
 * @return {number}
 */
TripleInOne.prototype.pop = function (stackNum) {
    if (this.stack[stackNum] && this.stack[stackNum].length) {
        return this.stack[stackNum].pop()
    }
    return -1

};

/** 
 * @param {number} stackNum
 * @return {number}
 */
TripleInOne.prototype.peek = function (stackNum) {
    if (this.stack[stackNum] && this.stack[stackNum].length) {
        return this.stack[stackNum][this.stack[stackNum].length - 1]
    }
    return -1

};

/** 
 * @param {number} stackNum
 * @return {boolean}
 */
TripleInOne.prototype.isEmpty = function (stackNum) {
    return !this.stack[stackNum] || !this.stack[stackNum].length
};

/**
 * Your TripleInOne object will be instantiated and called as such:
 * var obj = new TripleInOne(stackSize)
 * obj.push(stackNum,value)
 * var param_2 = obj.pop(stackNum)
 * var param_3 = obj.peek(stackNum)
 * var param_4 = obj.isEmpty(stackNum)
 */

来源：力扣（LeetCode）

链接：leetcode.cn/problems/th…

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