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; } };