LeetCode P260 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?

题目大意：给定一个整数数组，数组中除了两个数之外，其他数都是成对存在。请找出这两个数。

思路：该题思路类似SingNumber。假如要找的两个数为x、y，先将数组中所有数相互异或，得到的结果为result = x⊕y。result转换成二进制后一定存在值为1的位。找到result中任意一位值为1的位，然后按照这一位将数组中的数分为两个数组。一个数组是那一位是1的，另一个是那一位为0的。然后两个数组的数分别异或，最后的结果就是需要找的那两个数。

JavaCode：

public class Solution {    public int[] singleNumber(int[] nums) {    int result = 0;    //将数组中的数逐个异或        for (int i = 0; i < nums.length; i++) {result ^= nums[i];}        int bit = 1;        //查找出结果中任意值为1的位        while (true) {if ((result & bit) > 0) {break;}else {bit <<= 1;}}        int one = 0;        int two = 0;        //分成两组并各组逐一异或        for (int i = 0; i < nums.length; i++) {if ((nums[i] & bit) > 0) {one ^= nums[i];}else {two ^= nums[i];}}    return new int[]{one,two};    }}