【LeetCode】136. Single Number

• 2019 年 11 月 8 日
• 筆記

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本文链接：https://blog.csdn.net/shiliang97/article/details/102092957

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

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/single-number 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这道题用异或做太火了。。。。

class Solution {  public:      int singleNumber(vector<int>& nums) {          int num=0;          for(auto it :nums){              num ^= it;          }          return num;      }  };

想了一个，排序然后俩俩一对儿

class Solution {  public:      int singleNumber(vector<int>& nums) {          int n =nums.size();          if(n==1){              return nums[0];          }          sort(nums.begin(),nums.end());          int i=0;          while(i+1<n){              if(nums[i]!=nums[i+1]){                  return nums[i];              }              i=i+2;          }          return nums[i];      }  };