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题目：
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?
解析：
1 class Solution { 2 public: 3 int missingNumber(vector<int>& nums) { 4 sort(nums.begin(),nums.end()); 5 for(int i = 0; i <= nums.size(); i++) 6 if(nums[i] != i) 7 return i; 8 } 9 };
高效的算法应该是使用位运算，我没明白，也不想深究：
class Solution { public: int missingNumber(vector<int>& nums) { int result = 0; for (int i = 0; i < nums.size(); i++) result ^= nums[i]^(i+1); return result; } };
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原文地址：http://www.cnblogs.com/raichen/p/4952193.html
