Problem: 384. 打乱数组

题目描述

思路

打乱数组的主要算法：

从1 - n每次生成[i ~ n - i]的一个随机数字，再将原数组下标位置为i的元素和该随机数字位置的元素交换

复杂度

打乱数组的主要算法
时间复杂度:

$O(n)$；其中$n$为数组的大小

空间复杂度:

$O(1)$

Code

class Solution {
    private int[] nums;
    private Random rand = new Random();

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

    public int[] reset() {
        return nums;
    }

    /**
     * Shuffling algorithm
     *
     * @return int[]
     */
    public int[] shuffle() {
        int n = nums.length;
        int[] copy = Arrays.copyOf(nums, n);
        for (int i = 0; i < n; ++i) {
            // Generate a random number in the [i, n-1] range
            int r = i + rand.nextInt(n - i);
            // Exchange nums[i] and nums[r]
            swap(copy, i, r);
        }
        return copy;
    }

    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

/**
 * 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();
 */