268. Missing Number

Posted 2020-10-31 一日一更

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。

 

1. class Solution {  
2. public:  
3.     int missingNumber(vector<int>& nums) {  
4.         if(nums.size()==0) return 0;  
5.         nums.push_back(-1);  
6.         for(int i=0;i<nums.size();i++){  
7.             int temp;  
8.             int index=nums[i];  
9.             if(index==-1) break;  
10.               
11.             while(1){  
12.                 temp = nums[index];  
13.                 nums[index] = index;  
14.                 if(temp == nums[index] || temp ==-1) break;  
15.                 index = temp;  
16.             }  
17.               
18.         }  
19.          for(int i=0;i<=nums.size();i++)  
20.           if(nums[i]!=i) return i;  
21.           
22.     }  
23. };