[LeetCode]Missing Number

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Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

这题需要技巧，利用异或。

先将nums的数作异或得到x1；

将0-n+1都作异或得到x2；

missing number 就是x1^x2。

为什么呢？

假设缺少的值为A，则x2=x1^A，那么x1^x2=x1^(x1^A)=(x1^x1)^A=0^A=A，得证。

 1 class Solution {
 2 public:
 3     int missingNumber(vector<int>& nums) {
 4         int result;
 5         int x1=nums[0];
 6         for(int i=1;i<nums.size();i++)
 7         {
 8             x1^=nums[i];
 9         }
10         int x2=0;
11         for(int i=1;i<=nums.size();i++)
12         {
13             x2^=i;
14         }
15         result = x1^x2;
16         return result;
17     }
18 };

[LeetCode]Missing Number

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原文地址：http://www.cnblogs.com/Sean-le/p/4797911.html