LeetCode 384. Shuffle an Array

原题链接在这里：https://leetcode.com/problems/shuffle-an-array/

题目：

Shuffle a set of numbers without duplicates.

Example:

// Init an array with set 1, 2, and 3.
int[] nums = {1,2,3};
Solution solution = new Solution(nums);

// Shuffle the array [1,2,3] and return its result. Any permutation of [1,2,3] must equally likely to be returned.
solution.shuffle();

// Resets the array back to its original configuration [1,2,3].
solution.reset();

// Returns the random shuffling of array [1,2,3].
solution.shuffle();

题解：

Shuffle时每个数字在小于它的方位内 swap 一次.

j = random.nextInt(i+1). j is picked randomly from [0,i].

有两种可能

1. j == i, 那么nums[i]就没有换位置. 概率是1/(1+i).

2. j != i, 那么nums[i]就换位置了, 另外[0, i-1] 之间 i choices, 任选一个数与其对调. 概率是 (1-1/(1+i)) * (1/i) = 1/(1+i).

以此类推后就表明，结果中任何一个数字都有相等的概率在任意个位置上. 就是any permutation has same 概率.

Time Complexity: reset, O(1). shuffle, O(nums.length).

Space: O(nums.length).

AC Java:

class Solution {
    int [] nums;
    Random random;
    
    public Solution(int[] nums) {
        this.nums = nums;
        random = new Random();
    }
    
    /** Resets the array to its original configuration and return it. */
    public int[] reset() {
        return this.nums;
    }
    
    /** Returns a random shuffling of the array. */
    public int[] shuffle() {
        int [] arr = nums.clone();
        for(int i = 0; i<nums.length; i++){
            int j = random.nextInt(i+1);
            swap(arr, i, j);
        }
        
        return arr;
    }
    
    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();
 */