讨论一下1060. Missing Element in Sorted Array，互相加米

O（N）的好做。但是用二分法为啥下面这种会出错呢？少的输入可以通过，最后两个很大的vector就出错，差一。顺求互相加米，谢谢

1.     int missingElement(vector<int>& nums, int k) {
   
2.         int n=nums.size();
   
3.         if (n==0 || k==0)
   
4.            return k;
   
5.         int start=nums[0], res=start+k;
   
6.         
   
7.         auto upBnd=upper_bound(nums.begin(), nums.end(), res);
   
8.         int dist=upBnd-nums.begin();
   
9.         
   
10.         if (dist==1)
    
11.             return res;
    
12.         else
    
13.             return res+dist-1;
    
14.     }

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1060. Missing Element in Sorted Array
Given a sorted array A of unique numbers, find the K-th missing number starting from the leftmost number of the array.



Example 1:

Input: A = [4,7,9,10], K = 1
Output: 5
Explanation:
The first missing number is 5.
Example 2:

Input: A = [4,7,9,10], K = 3
Output: 8
Explanation:
The missing numbers are [5,6,8,...], hence the third missing number is 8.
Example 3:

Input: A = [1,2,4], K = 3
Output: 6
Explanation:
The missing numbers are [3,5,6,7,...], hence the third missing number is 6.


Note:

1 <= A.length <= 50000
1 <= A <= 1e7
1 <= K <= 1e8

用java的看不懂