[leetcode]384. Shuffle an Array

Proof: Suppose this algorithm works, i.e. for each position j (starting from i), the probability of any number in the range[i,n-1] to be at position j is 1/(n-i).

Let’s look at int i = random.nextInt(i, n):
(1) If j == i, nums[i] does not need to change its position, which has probability 1/(n-i).
(2) if j != i, nums[i] needs to be swapped with nums[j]. The probability of any number x in the range [i, n-1] to be at position j = nums[j] changes its position * x is at position j
= (1-1/(n-i)) * (1/n-i-1) = 1/(n-i) （选的位置是要和i交换的概率 * number x正好是选中的位置的概率）

Each number has equal probability to be at any position.

class Solution {
    private int[] array;
    private int[] original;
    private 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();
    }
    
    /** Resets the array to its original configuration and return it. */
    public int[] reset() {
        array = original;
        original = original.clone();
        return original;
    }
    
    /** Returns a random shuffling of the array. */
    public int[] shuffle() {
        for(int i = 0;i < original.length;i++){
            swapAt(i, randRange(i, original.length));
        }
        
        return array;
    }
}

/**
 * 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();
 */