一、题目

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。

点击此处跳转题目。

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]

提示：

0 <= stackNum <= 2

二、C# 题解

很基础的题目，代码如下：

public class TripleInOne {private int[][] stack; // 2 维数组存储 3 个栈private int[] p;       // 1 维数组存储 3 个栈的指针public TripleInOne(int stackSize) {stack = new int[3][];for (int i = 0; i < 3; i++) {stack[i] = new int[stackSize];}p = new int[] {-1, -1, -1};}public void Push(int stackNum, int value) {if (p[stackNum] == stack[stackNum].Length - 1) return;stack[stackNum][++p[stackNum]] = value;}public int Pop(int stackNum) {if (p[stackNum] == -1) return -1;return stack[stackNum][p[stackNum]--];}public int Peek(int stackNum) {if (p[stackNum] == -1) return -1;return stack[stackNum][p[stackNum]];}public bool IsEmpty(int stackNum) {return p[stackNum] == -1;}
}/*** Your TripleInOne object will be instantiated and called as such:* TripleInOne obj = new TripleInOne(stackSize);* obj.Push(stackNum,value);* int param_2 = obj.Pop(stackNum);* int param_3 = obj.Peek(stackNum);* bool param_4 = obj.IsEmpty(stackNum);*/