Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,1] Output: 1
Example 2:
Input: [4,1,2,1,2] Output: 4
这道题我本来想的是用find函数查找,若有两个数一样就erase删除。但是最后看到这个超级简单的方法。
代码:
class Solution { public: int singleNumber(vector<int>& nums) { int val=0; for(int i=0; i<nums.size(); i++) { val^=nums[i]; } return val; } };
代码2:(排序)
这个要比第一个慢一些,但也是一种方法,所以也放在这里了~
public: int singleNumber(vector<int>& nums) { int i; if(nums.size()==1) return nums[0]; sort(nums.begin(), nums.end()); for(i=0; i<nums.size()-1; i=i+2) { if(nums[i]!=nums[i+1]) return nums[i]; } return nums[i]; } };