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]

提示：

0 <= stackNum <= 2

class TripleInOne {
	int s[99999];
	int sSize = 0;
	int size[3];
	const int STACK_SIZE = 3;
public:
	TripleInOne(int stackSize) {
		sSize = stackSize;
		size[0] = 0;
		size[1] = 1;
		size[2] = 2;
	}

	void push(int stackNum, int value) {
		if (size[stackNum]>=(STACK_SIZE)*sSize)
		{
			return;
		}
		s[size[stackNum]] = value;
		size[stackNum] += STACK_SIZE;
	}

	int pop(int stackNum) {
		if (isEmpty(stackNum))
		{
			return -1;
		}
        int result=peek(stackNum);
		size[stackNum] -= STACK_SIZE;
        return result;
	}

	int peek(int stackNum) {
		if (isEmpty(stackNum))
		{
			return -1;
		}
		return s[size[stackNum] - STACK_SIZE];
	}

	bool isEmpty(int stackNum) {
		return size[stackNum] - stackNum <= 0;
	}
};

/**
 * 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);
 */