136.Single Number

相关链接

leetcode

背景知识

Concept原理

• If we take XOR of zero and some bit, it will return that bit
  a⊕0=a（0异或任何数得任何数）

• If we take XOR of two same bits, it will return 0
  a⊕a=0（相同的两个数异或为0）

• a⊕b⊕a=(a⊕a)⊕b=0⊕b=b（异或满足交换律）

So we can XOR all bits together to find the unique number.

描述

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?

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Example 1:

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

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

solution

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int len = nums.size();
        for(int i = 1; i < len; i++)
            nums[0] ^= nums[i];           
        return nums[0];
    }
};

leetcode 136.Single Number

原文：https://www.cnblogs.com/qwfand/p/12615072.html