LeetCode: SingleNumIII

題目：

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? 題意：給定一個數組nums，其中除了兩個元素只出現一次外，其他元素均出現兩次，找出僅出現一次的兩個元素 提示：1.對輸出結果的順序不作要求 2.算法要求線性復雜度 分析：仍然是異或的用法，題目較簡單，直接看代碼，代碼中有注釋 代碼：

   public class Solution {
       public int[] singleNumber(int[] nums) {
           int[] single = new int[2];
           int[] wei = new int[32];
           int yhResult = 0;
           
           for(int i=0; i				//如果異或結果的某位為1，說明兩個單數在該位是不同的
                   for(int j=0; j				//按該位相同與不同，分兩組分別異或，可直接得出結果元素
                           single[0] ^= nums[j];
                       else
                           single[1] ^= nums[j];
                   }
                   break;
               }
           }
            return single;
       }
      
   }