力扣每日一题2021/8/18

题目：384. 打乱数组

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。

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

难度：中等

示例：

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 <= 200
• -10^6 <= nums[i] <= 10^6
• nums 中的所有元素都是 唯一的
• 最多可以调用 5 * 10^4 次 reset 和 shuffle

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/shuffle-an-array
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

解题思路

1. 暴力
2. Fisher-Yates洗牌算法

解题代码

暴力

class Solution {

    public int[] nums;
    public Solution(int[] nums) {
        this.nums = nums;
    }

    /** Resets the array to its original configuration and return it. */
    public int[] reset() {
        return nums;
    }

    /** Returns a random shuffling of the array. */
    public int[] shuffle() {
        int[] res = new int[nums.length];
        HashSet<Integer> hashSet = new HashSet<>();
        Random random = new Random();
        for (int i = 0; i < res.length; i++){
            int rand = random.nextInt(res.length);
            if (hashSet.add(rand)){
                res[i] = nums[rand];
            }else {
                i--;
            }
        }
        return res;
    }
}

/**
 * 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();
 */

官方解题代码

暴力

class Solution {
    private int[] array;
    private int[] original;

    private Random rand = new Random();

    private List<Integer> getArrayCopy() {
        List<Integer> asList = new ArrayList<Integer>();
        for (int i = 0; i < array.length; i++) {
            asList.add(array[i]);
        }
        return asList;
    }

    public Solution(int[] nums) {
        array = nums;
        original = nums.clone();
    }
    
    public int[] reset() {
        array = original;
        original = original.clone();
        return array;
    }
    
    public int[] shuffle() {
        List<Integer> aux = getArrayCopy();

        for (int i = 0; i < array.length; i++) {
            int removeIdx = rand.nextInt(aux.size());
            array[i] = aux.get(removeIdx);
            aux.remove(removeIdx);
        }

        return array;
    }
}

Fisher-Yates洗牌算法

class Solution {
    private int[] array;
    private int[] original;

    Random rand = new Random();

    private int randRange(int min, int max) {
        return rand.nextInt(max - min) + min;
    }

    private void swapAt(int i, int j) {
        int temp = array[i];
        array[i] = array[j];
        array[j] = temp;
    }

    public Solution(int[] nums) {
        array = nums;
        original = nums.clone();
    }
    
    public int[] reset() {
        array = original;
        original = original.clone();
        return original;
    }
    
    public int[] shuffle() {
        for (int i = 0; i < array.length; i++) {
            swapAt(i, randRange(i, array.length));
        }
        return array;
    }
}