Java [Leetcode 384]Shuffle an Array

題目描述：

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();

解題思路：

每次往后讀取數組的時候，當讀到第i個的時候以1/i的概率隨機替換1～i中的任何一個數，這樣保證最后每個數字出現在每個位置上的概率都是相等的。

證明：

設x元素在第m次的時候出現在位置i的概率是1/m,那么在第m+1次的時候，x仍然待在位置i的概率是 1/m * m/(m+1) = 1/(m+1)

代碼描述：

public class Solution {

	private int[] nums = null;
	private Random random = null;

    public Solution(int[] nums) {
        this.nums = nums;
        random = new Random();
    }
    
    /** 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() {
        if(nums == null) return null;
        int[] a = (int[])nums.clone();
        for(int i = 1; i < nums.length; i++){
        	int j = random.nextInt(i + 1);
        	swap(a, i, j);
        }
        return a;
    }

    private void swap(int[] a, int i, int j){
    	int temp = a[i];
    	a[i] = a[j];
    	a[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();
 */