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136. Single Number

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: 1Example 2:Input: [4,1,2,1,2]Output: 4

Solution

Bit Operation

Runtime: 24 ms, faster than 30.74% of C++ online submissions for Single Number.

class Solution {   public:    int singleNumber(vector
   
    & nums) {           int a = 0;        for(int i = 0; i < nums.size(); i++)        {               a^=nums[i];        }        return a;    }};
   

这是个比较巧妙的方法，采用位运算——异或。根据异或的性质有：

Hash Table

Runtime: 36 ms, faster than 16.05% of C++ online submissions for Single Number.

class Solution{       public:        int singleNumber(vector
   
    & nums){               unordered_map 
    
      nums_map;            for(int i = 0; i < nums.size(); i++)                nums_map[nums[i]]++;            for(auto it = nums_map.begin(); it != nums_map.end(); it++)                if(it->second == 1) return it->first;            return nums[0];        }};
    
   

it迭代器：

unordered_map
   
    ::iterator it;(*it).first;             // the key value (of type Key)(*it).second;            // the mapped value (of type T)(*it);                   // the "element value" (of type pair
    
     )
    
   

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