260. Single Number III

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

public class Solution {
    public int[] singleNumber(int[] nums) {
        int result[] = {0,0};
        if(nums == null || nums.length == 0)
            return result;
        int tmp = nums[0];
        for(int i = 1;i< nums.length; i++){
            tmp ^= nums[i];
        }
        int firstone = tmp & (-tmp);
        for(int j = 0;j< nums.length;j++){
            if((nums[j] & firstone) != 0){//位运算级别比较低，需要括号
                result[0] ^= nums[j];
            }else{
                result[1] ^= nums[j];
            }
        }
        return result;
    }
}

int firstone = tmp & (-tmp); 相与找出从右往左第一个为1的位置，再将数组分为两组，其中一组数字该位为1，另一组该位为0.
剩下的解法和 Single Number问题相似。

260. Single Number III

原文：http://www.cnblogs.com/guoguolan/p/5392584.html