Leetcode 2200. Find all k nearest neighbor subscripts in the array (yes, one pass)

I'll give you a subscript from 0 The starting array of integers nums And two integers key and k .K Nearest neighbor subscript yes nums A subscript in i , And satisfy that there is at least one subscript j bring |i - j| <= k And nums[j] == key .

To return as a list, press Increasing order Sort all of K Nearest neighbor subscript .

Example 1：

 Input ：nums = [3,4,9,1,3,9,5], key = 9, k = 1
 Output ：[1,2,3,4,5,6]
 explain ： therefore ,nums[2] == key  And  nums[5] == key .
-  Subscript to subscript  0 ,|0 - 2| > k  And  |0 - 5| > k , So there is no  j  bring  |0 - j| <= k  And  nums[j] == key . therefore  0  Is not a  K  Nearest neighbor subscript .
-  Subscript to subscript  1 ,|1 - 2| <= k  And  nums[2] == key , therefore  1  It's a  K  Nearest neighbor subscript .
-  Subscript to subscript  2 ,|2 - 2| <= k  And  nums[2] == key , therefore  2  It's a  K  Nearest neighbor subscript .
-  Subscript to subscript  3 ,|3 - 2| <= k  And  nums[2] == key , therefore  3  It's a  K  Nearest neighbor subscript .
-  Subscript to subscript  4 ,|4 - 5| <= k  And  nums[5] == key , therefore  4  It's a  K  Nearest neighbor subscript .
-  Subscript to subscript  5 ,|5 - 5| <= k  And  nums[5] == key , therefore  5  It's a  K  Nearest neighbor subscript .
-  Subscript to subscript  6 ,|6 - 5| <= k  And  nums[5] == key , therefore  6  It's a  K  Nearest neighbor subscript .
 therefore , Returns... In ascending order  [1,2,3,4,5,6] . 

Example 2：

 Input ：nums = [2,2,2,2,2], key = 2, k = 2
 Output ：[0,1,2,3,4]
 explain ： Yes  nums  All subscripts of  i , There is always a subscript  j  bring  |i - j| <= k  And  nums[j] == key , So every subscript is a  K  Nearest neighbor subscript . 
 therefore , return  [0,1,2,3,4] .

Tips ：

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 1000
• key It's an array nums An integer in
• 1 <= k <= nums.length

Code:

class Solution {
    
public:
    vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
    
        vector<int>pos;
        for(int i=0;i<nums.size();i++)
        {
    
            if(nums[i]==key)
            {
    
                pos.push_back(i);
            }
        }
        vector<int>res;
        for(int i=0;i<nums.size();i++)
        {
    
            for(int j=0;j<pos.size();j++)
            {
    
                if(abs(i-pos[j])<=k)
                {
    
                    res.push_back(i);
                    break;
                }
            }
        }
        sort(res.begin(),res.end());
        return res;
    }
};