#137 Single Number II

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

解法：先对数组进行排序， 然后进行统计出现的个数，如果等于3，则进行循环，否则就跳出循环了

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int len = nums.size();
        sort(nums.begin(), nums.end());
        int start = nums[0] ;
        int count = 1;
        for(int i = 1; i < len; i++) {
            if(start == nums[i]) {
                count++;
            }
            else {
                if(count != 3) return start;
                else {
                start = nums[i];
                count = 1;
                }
            }
        }
        return start;  
    }
};