本文共 1909 字，大约阅读时间需要 6 分钟。
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
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 ⊕ a = 0 , a ⊕ 0 = a , a ⊕ a ⊕ b = b , ( a ⊕ b ) ⊕ ( a ⊕ b ) = ( a ⊕ a ) ⊕ ( b ⊕ b ) = 0 \\ 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 a⊕a=0,a⊕0=a,a⊕a⊕b=b,(a⊕b)⊕(a⊕b)=(a⊕a)⊕(b⊕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。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：