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LeetCode

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Problem

  • 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

Code

class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        int left = 0;
        int right = arr.length - k;

        while (left < right) {
            int mid = (left + right) / 2;

            if (x < arr[mid]) {
                right = mid;
            } else if (x > arr[mid + k]) {
                left = mid + 1;
            } else {
                if (x - arr[mid] <= arr[mid + k] - x) {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }
        }

        List<Integer> res = new ArrayList<>(k);
        for (int i = left; i < left + k; i++) {
            res.add(arr[i]);
        }

        return res;
    }
}