268. Missing Number

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

Example 1

Input: [3,0,1]
Output: 2

Example 2

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

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.

解题思路：

在最后添加一个-1值，对数组从0-N对每个数字进行归位，最后数字不符合的就是丢失的那个数字。

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        if(nums.size()==0) return 0;
        nums.push_back(-1);
        for(int i=0;i<nums.size();i++){
            int temp;
            int index=nums[i];
            if(index==-1) break;

            while(1){
                temp = nums[index];
                nums[index] = index;
                if(temp == nums[index] || temp ==-1) break;
                index = temp;
            }

        }
         for(int i=0;i<=nums.size();i++)
          if(nums[i]!=i) return i;

    }
};