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  • 658. Find K Closest Elements

    Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.

    Example 1:

    Input: [1,2,3,4,5], k=4, x=3
    Output: [1,2,3,4] 

    Example 2:

    Input: [1,2,3,4,5], k=4, x=-1
    Output: [1,2,3,4] 

    Note:

    1. The value k is positive and will always be smaller than the length of the sorted array.
    2. Length of the given array is positive and will not exceed 104
    3. Absolute value of elements in the array and x will not exceed 104

    Solution 1: set a map, the key is |arr[i]-x|, the value is the index vector, eg, 1:[1,3] for example 1 input.  O(klogk+n)

     1 class Solution {
     2 public:
     3     vector<int> findClosestElements(vector<int>& arr, int k, int x) {
     4         map<int,vector<int>> m;
     5         vector<int> res;
     6         for (int i=0;i<arr.size();i++){
     7             int distance=abs(arr[i]-x);
     8             m[distance].push_back(i);
     9         }
    10         int count=0;
    11         for (auto it=m.begin();it!=m.end();it++){
    12                 int size=it->second.size();
    13                 for (int i=0;i<size;i++){
    14                     if (count<k){
    15                         int index=it->second[i];
    16                         res.push_back(arr[index]);
    17                         count++;
    18                     }                    
    19                 }
    20                                     
    21         }
    22         sort(res.begin(),res.end());
    23         return res;
    24     }
    25 };

    Solution 2: use the binary search to find the first number which is >=x, then, determine the indices of the start and the end of a subarray in arr, where the subarray is our result. The time complexity is O(logn + k).

     1 class Solution {
     2 public:
     3     vector<int> findClosestElements(vector<int>& arr, int k, int x) {
     4         int index=lower_bound(arr.begin(),arr.end(),x)-arr.begin(); //binary search:log(n)
     5         int i=index-1,j=index;
     6         while(k--){
     7             if (i<0||(j<arr.size()&&abs(arr[j]-x)<abs(arr[i]-x))) j++;
     8             else i--;
     9         }
    10         vector<int> res(arr.begin()+i+1,arr.begin()+j);
    11         return res;
    12     }
    13 };

     line 4 use the lower_bound which is a binary search function like below.

    int binarySearch(vector<int>& arr,int x){
            int beg=0,end=arr.size()-1,pos;
            while (beg<end){
                int mid=(beg+end)/2;
                if (arr[mid]==x) return mid;
                if (arr[mid]<x) {beg=mid+1;pos=beg;}
                if (arr[mid]>x) {end=mid;pos=end;}
            }
            return pos;
        }
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  • 原文地址:https://www.cnblogs.com/anghostcici/p/7359833.html
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