658. Find K Closest Elements

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本文介绍了一种使用优先队列寻找给定数组中最接近指定值的k个元素的方法，并通过一个Java实现的例子展示了如何确保结果按升序排列。该算法适用于长度不超过10^4的已排序数组。

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:

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

java

class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        Comparator<Integer> cmp = new Comparator<Integer>() {
            public int compare(Integer a, Integer b) {
                int d1 = Math.abs(a - x);
                int d2 = Math.abs(b - x);
                int diff = d2 - d1;
                return diff;
            }
        };
    Queue<Integer> queue = new PriorityQueue<>(k, cmp);
    for (int i = 0; i < arr.length; i++) {
        if (queue.size() < k) {
            queue.offer(arr[i]);
        } else {
            int temp = queue.peek();
            int v1 = Math.abs(temp - x);
            int v2 = Math.abs(arr[i] - x);
            if (v1 > v2) {
                queue.poll();
                queue.offer(arr[i]);
            }
        }
    }
    List<Integer> list = new ArrayList<>();
    while(!queue.isEmpty()) {
        list.add(queue.poll());
    }
    Collections.sort(list);
    return list;
}
}