打乱数组

• https://leetcode.cn/problems/shuffle-an-array/description/

描述

• 给你一个整数数组 nums ，设计算法来打乱一个没有重复元素的数组
• 打乱后，数组的所有排列应该是 等可能 的
• 实现 Solution class:
  • Solution(int[] nums) 使用整数数组 nums 初始化对象
  • int[] reset() 重设数组到它的初始状态并返回
  • int[] shuffle() 返回数组随机打乱后的结果

示例 1

输入
["Solution", "shuffle", "reset", "shuffle"]
[[[1, 2, 3]], [], [], []]
输出
[null, [3, 1, 2], [1, 2, 3], [1, 3, 2]]

解释
Solution solution = new Solution([1, 2, 3]);
solution.shuffle(); // 打乱数组 [1,2,3] 并返回结果。任何 [1,2,3]的排列返回的概率应该相同。例如，返回 [3, 1, 2]
solution.reset(); // 重设数组到它的初始状态 [1, 2, 3] 。返回 [1, 2, 3]
solution.shuffle(); // 随机返回数组 [1, 2, 3] 打乱后的结果。例如，返回 [1, 3, 2]

提示

• 1 <= nums.length <= 50
• -$10^6$ <= nums[i] <= $10^6$
• nums 中的所有元素都是 唯一的
• 最多可以调用 104 次 reset 和 shuffle

Typescript 版算法实现

1 ) 方案1：暴力

class Solution {
    private nums: number[];
    private original: number[];

    constructor(nums: number[]) {
        this.nums = [...nums];
        this.original = [...nums];
    }

    reset(): number[] {
        this.nums = [...this.original];
        return this.nums;
    }

    shuffle(): number[] {
        const shuffled = [...this.nums]; // 创建一个副本避免直接修改原数组
        for (let i = shuffled.length - 1; i > 0; i--) {
            const j = Math.floor(Math.random() * (i + 1));
            [shuffled[i], shuffled[j]] = [shuffled[j], shuffled[i]]; // Fisher-Yates 洗牌算法
        }
        this.nums = shuffled;
        return this.nums;
    }
}


/**
 * Your Solution object will be instantiated and called as such:
 * var obj = new Solution(nums)
 * var param_1 = obj.reset()
 * var param_2 = obj.shuffle()
 */

2 ) 方案2：Fisher-Yates 洗牌算法

class Solution {
    private nums: number[];
    private original: number[];

    constructor(nums: number[]) {
        this.nums = [...nums];
        this.original = [...nums];
    }

    reset(): number[] {
        this.nums = [...this.original];
        return this.nums;
    }

    shuffle(): number[] {
        const shuffled = [...this.nums]; // 创建一个副本避免直接修改原数组
        for (let i = 0; i < shuffled.length; ++i) {
            const j = Math.floor(Math.random() * (shuffled.length - i)) + i;
            [shuffled[i], shuffled[j]] = [shuffled[j], shuffled[i]]; // 解构赋值交换元素
        }
        this.nums = shuffled;
        return this.nums;
    }
}


/**
 * Your Solution object will be instantiated and called as such:
 * var obj = new Solution(nums)
 * var param_1 = obj.reset()
 * var param_2 = obj.shuffle()
 */