Add to List 658. Find K Closest Elements

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本文介绍了一种算法，用于从已排序数组中找到指定数量的最接近目标值的元素，并确保结果按升序排列。该算法考虑了边界情况，并通过二分查找定位起始位置，再向两边扩展来选取最佳元素。

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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

class Solution {
    public List<Integer> findClosestElements(List<Integer> arr, int k, int x) {
        if (x <= arr.get(0))
			return arr.subList(0, k);
		if (x >= arr.get(arr.size() - 1))
			return arr.subList(arr.size() - k, arr.size());
		LinkedList<Integer> re = new LinkedList<>();
		int lo = 0;
		int hi = arr.size() - 1;
		int i = -1;
		while (lo <= hi) {
			int mid = lo + (hi - lo) / 2;
			if (arr.get(mid) == x) {
				i = mid;
				break;
			} else if (arr.get(mid) < x)
				lo = mid + 1;
			else
				hi = mid - 1;
		}
		if (i == -1) {
			int lo_gap = Math.abs(arr.get(lo) - x);
			int hi_gap = Math.abs(arr.get(hi) - x);
			i = lo_gap > hi_gap ? hi : lo;
		}
		re.add(arr.get(i));
		lo = i - 1;
		hi = i + 1;
		while (--k > 0) {
			int lo_gap = Integer.MAX_VALUE;
			int hi_gap = Integer.MAX_VALUE;
			if (lo >= 0)
				lo_gap = Math.abs(arr.get(lo) - x);
			if (hi < arr.size())
				hi_gap = Math.abs(arr.get(hi) - x);
			if (lo_gap <= hi_gap) {
				re.addFirst(arr.get(lo));
				lo--;
			} else {
				re.addLast(arr.get(hi));
				hi++;
			}
		}
		return re;
    }
}