LeetCode Find K Closest Elements

标签：desc   targe   integer   else   题目   ++   inpu   log   osi   

原题链接在这里：https://leetcode.com/problems/find-k-closest-elements/description/

题目：

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

题解：

直接查找这k个最近值开始的位置po. 也就是arr[po] 比 arr[po+k]更接近目标值x.

Time Complexity: O(logn + k).

Space: O(1). regardless res.

AC Java:

 1 class Solution {
 2     public List<Integer> findClosestElements(int[] arr, int k, int x) {
 3         List<Integer> res = new ArrayList<Integer>();
 4         if(arr == null || arr.length == 0 || k == 0){
 5             return res;
 6         }
 7         
 8         int l = 0;
 9         int r = arr.length-k;
10         while(l<r){
11             int mid = l+(r-l)/2;
12             if(x - arr[mid] > arr[mid+k]-x){
13                 l = mid+1;
14             }else{
15                 r = mid;
16             }
17         }
18         
19         for(int i = l; i<l+k; i++){
20             res.add(arr[i]);
21         }
22         return res;
23     }
24 }

LeetCode Find K Closest Elements

标签：desc   targe   integer   else   题目   ++   inpu   log   osi   

原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/7949934.html