README.md

面试题 03.01. 三合一

English Version

题目描述

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]

解法

二维数组解决；也可以使用一维数组，以下标 0,3,6,..、1,4,7,..、2,5,8,.. 区分，一维数组最后三个元素记录每个栈的元素个数。

Python3

class TripleInOne:

    def __init__(self, stackSize: int):
        self._capacity = stackSize
        self._s = [[], [], []]

    def push(self, stackNum: int, value: int) -> None:
        if len(self._s[stackNum]) < self._capacity:
            self._s[stackNum].append(value)

    def pop(self, stackNum: int) -> int:
        return -1 if self.isEmpty(stackNum) else self._s[stackNum].pop()

    def peek(self, stackNum: int) -> int:
        return -1 if self.isEmpty(stackNum) else self._s[stackNum][-1]

    def isEmpty(self, stackNum: int) -> bool:
        return len(self._s[stackNum]) == 0


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

Java

class TripleInOne {
    private int[] s;
    private int capacity;

    public TripleInOne(int stackSize) {
        s = new int[stackSize * 3 + 3];
        capacity = stackSize;
    }

    public void push(int stackNum, int value) {
        if (s[stackNum + 3 * capacity] < capacity) {
            s[s[stackNum + 3 * capacity] * 3 + stackNum] = value;
            ++s[stackNum + 3 * capacity];
        }
    }

    public int pop(int stackNum) {
        if (isEmpty(stackNum)) {
            return -1;
        }
        --s[stackNum + 3 * capacity];
        return s[s[stackNum + 3 * capacity] * 3 + stackNum];
    }

    public int peek(int stackNum) {
        return isEmpty(stackNum) ? -1 : s[(s[stackNum + 3 * capacity] - 1) * 3 + stackNum];
    }

    public boolean isEmpty(int stackNum) {
        return s[stackNum + 3 * capacity] == 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);
 * boolean param_4 = obj.isEmpty(stackNum);
 */

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