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面试题 03.03. 堆盘子

English Version

题目描述

堆盘子。设想有一堆盘子，堆太高可能会倒下来。因此，在现实生活中，盘子堆到一定高度时，我们就会另外堆一堆盘子。请实现数据结构SetOfStacks，模拟这种行为。SetOfStacks应该由多个栈组成，并且在前一个栈填满时新建一个栈。此外，SetOfStacks.push()和SetOfStacks.pop()应该与普通栈的操作方法相同（也就是说，pop()返回的值，应该跟只有一个栈时的情况一样）。 进阶：实现一个popAt(int index)方法，根据指定的子栈，执行pop操作。

当某个栈为空时，应当删除该栈。当栈中没有元素或不存在该栈时，pop，popAt 应返回 -1.

示例1:

 输入：
["StackOfPlates", "push", "push", "popAt", "pop", "pop"]
[[1], [1], [2], [1], [], []]
 输出：
[null, null, null, 2, 1, -1]

示例2:

 输入：
["StackOfPlates", "push", "push", "push", "popAt", "popAt", "popAt"]
[[2], [1], [2], [3], [0], [0], [0]]
 输出：
[null, null, null, null, 2, 1, 3]

解法

Python3

Java

TypeScript

class StackOfPlates {
    private cap: number;
    private stacks: number[][];

    constructor(cap: number) {
        this.cap = cap;
        this.stacks = [];
    }

    push(val: number): void {
        if (this.cap === 0) {
            return;
        }
        const n = this.stacks.length;
        const stack = this.stacks[n - 1];
        if (stack == null || stack.length === this.cap) {
            this.stacks.push([val]);
        } else {
            stack.push(val);
        }
    }

    pop(): number {
        const n = this.stacks.length;
        if (n === 0) {
            return -1;
        }
        const stack = this.stacks[n - 1];
        const res = stack.pop();
        if (stack.length === 0) {
            this.stacks.pop();
        }
        return res;
    }

    popAt(index: number): number {
        if (index >= this.stacks.length) {
            return -1;
        }
        const stack = this.stacks[index];
        const res = stack.pop();
        if (stack.length === 0) {
            this.stacks.splice(index, 1);
        }
        return res;
    }
}

/**
 * Your StackOfPlates object will be instantiated and called as such:
 * var obj = new StackOfPlates(cap)
 * obj.push(val)
 * var param_2 = obj.pop()
 * var param_3 = obj.popAt(index)
 */

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