LeetCode 658. Find K Closest Elements

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.

二分查找时，如果arr[po] 比 arr[po+k] 更加远离目标值，那么起始位置po肯定在当下试验位置的右侧.

Time Complexity: O(logn + k).

Space: O(1). regardless res.

AC Java:

class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        List<Integer> res = new ArrayList<Integer>();
        if(arr == null || arr.length == 0 || k == 0){
            return res;
        }
        
        int l = 0;
        int r = arr.length-k;
        while(l<r){
            int mid = l+(r-l)/2;
            if(x - arr[mid] > arr[mid+k]-x){
                l = mid+1;
            }else{
                r = mid;
            }
        }
        
        for(int i = l; i<l+k; i++){
            res.add(arr[i]);
        }
        return res;
    }
}