Single Number III leetcode java

问题描述：

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

提示：bit manipulation

一、位运算

//参照single number 的方法，将所有数异或运算，得到result为两个single number的异或；
//找到result中最低位为1的那一位，是两个single number不同的地方，根据这个bit，可以将nums数组中的数分为A、B两组；
//然后分别在A和B中寻找single number即可

public static int[] singleNumber_Bit(int[] nums){

    int result = 0;
    for(int i = 0; i < nums.length; i++){
        result = result ^ nums[i];
    }
    int[] res = new int[2];
    //找到result二进制表示中最右侧的1
    int pos = result & ( ~ (result - 1 )); //统计一个int型整数的二进制表示中有多少个1，可以采用 n = n & (n - 1);来从右到左逐个统计1的个数
    for(int i = 0 ; i < nums.length; i++){ //将nums分为A B两组来分别求single number
         if((pos & nums[i]) != 0){
                res[0] = res[0] ^ nums[i];
         } else {
                res[1] = res[1] ^ nums[i];
         }
    }
    return res;
}