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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?
思路：先排序再查找。
代码：
1 class Solution { 2 public: 3 int missingNumber(vector<int>& nums) { 4 int i=0,res=1; 5 sort(nums.begin(),nums.end()); 6 while(i<nums.size()&&(res&1)^(nums[i]&1)){ 7 res=nums[i++]; 8 } 9 return i<nums.size()?nums[i]-1:i; 10 } 11 };
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原文地址：http://www.cnblogs.com/Deribs4/p/5721412.html
