题目链接：384. 打乱数组 - 力扣（LeetCode）

先从一道没什么技术含量的题开始了。本题引入了Fisher Yates洗牌算法原理，其实质是相同概率下的元素选取和交换：从后向前（或从前向后）遍历数组，将当前遍历到的元素与其前面（后面）的任意数字交换后固定不变。取随机数的过程调用了java的Random类。

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// java解法——从后向前
import java.util.*;

class Solution {
    
    private int[] ori;
    private int[] nums;
    public Solution(int[] nums) {
        this.nums = nums;
        this.ori = nums.clone();
    }
    
    /** Resets the array to its original configuration and return it. */
    public int[] reset() {
        return this.ori;
    }
    
    /** Returns a random shuffling of the array. */
    public int[] shuffle() {
        Random random = new Random();
        int length = nums.length;
        for(int i = length - 1; i > 1; i--) {
            int j = random.nextInt(i + 1);
            int temp = nums[j];
            nums[j] = nums[i];
            nums[i] = temp;
        }
        return nums;
    }
}

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

python的话，import random也可以达到相同效果：

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def shuffle(self):
    for i in range(len(self.output)):
        j = random.randint(i, len(self.output) - 1)
        self.output[i], self.output[j] = self.output[j], self.output[i]
    return self.output