LeetCode 260. Single Number III （数组查重）

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:

The order of the result is not important. So in the above example, 

[5, 3]

 is
also correct.
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

输入一组整数，找出其中只出现一次的数，其余数都出现两次。

本题和136题（http://blog.csdn.net/hiroshiten/article/details/72637489）的区别在于，136题有且只有一个单独的数，本题可以有零个到多个。

思路：将数组排序，然后比较相邻的两个数是否相同。注意只有两个数且两个数不相同的情况。

vector<int> singleNumber(vector<int>& nums) {
if(nums.size()==2&&nums[0]!=nums[1])return nums;
sort(nums.begin(),nums.end());
vector<int> ans;
for(int i=0;i<nums.size();i++)
{
if(nums[i]!=nums[i+1])
ans.push_back(nums[i]);
else i++;
}
return ans;
}