【LeetCode】136. Single Number

• 2019 年 11 月 8 日
• 筆記

版權聲明：本文為博主原創文章，遵循 CC 4.0 BY-SA 版權協議，轉載請附上原文出處鏈接和本聲明。

本文鏈接：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

來源：力扣（LeetCode） 鏈接：https://leetcode-cn.com/problems/single-number 著作權歸領扣網絡所有。商業轉載請聯繫官方授權，非商業轉載請註明出處。

這道題用異或做太火了。。。。

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

想了一個，排序然後倆倆一對兒

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];      }  };