658. Find K Closest Elements

Given a sorted integer array arr, two integers k and x, return the k closest integers to x in the array. The result should also be sorted in ascending order.

An integer a is closer to x than an integer b if:

• |a - x| < |b - x|, or
• |a - x| == |b - x| and a < b

Example 1:

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

Example 2:

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

Constraints:

• 1 <= k <= arr.length
• 1 <= arr.length <= 104
• arr is sorted in ascending order.
• -104 <= arr[i], x <= 104

class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        List<Integer> list = new ArrayList();
        for(int i : arr) list.add(i);
        Collections.sort(list, (a, b) -> Math.abs(a - x) - Math.abs(b - x));
        List<Integer> res = list.subList(0, k);
        Collections.sort(res);
        return res;
    }
}

! 就摁用API，先按要求把list排好序。。属实妹想到，然后要多少拿多少，再排一次序。

class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        int lo = 0, hi = arr.length - 1;
        while(hi - lo >= k) {
            if(Math.abs(arr[lo] - x) > Math.abs(arr[hi] - x)) lo++;
            else hi--;
        }
        List<Integer> res = new ArrayList();
        for(int i = lo; i <= hi; i++) res.add(arr[i]);
        return res;
    }
}

用two pointers可以达到O（n）， 最终lo到hi就是答案