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.

Memory Usage: 12.1 MB, less than 6.17% of C++ online submissions for Single Number.

时间复杂度O(n),空间复杂度O(1)

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：

\\ a\oplus a=0, \\ a\oplus 0 = a, \\ a\oplus a\oplus b = b, \\ (a\oplus b) \oplus (a\oplus b) = (a\oplus a) \oplus (b\oplus b) = 0

所以，可以总结为：如果相同则对应位为0，如果不同则对应位为1。我们想要找到一个与众不同的元素，也就是需要每个元素间互相异或，这样所有相同的元素异或均为0，那么剩下的就是我们想要的答案了。

Hash Table

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

Memory Usage: 14 MB, less than 6.17% of C++ online submissions for Single Number.

时间复杂度O(n),空间复杂度O(n)

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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