K Closest Numbers In Sorted Array

寻找最接近的K个数

最新推荐文章于 2023-09-23 12:20:33 发布

weixin_34199335

最新推荐文章于 2023-09-23 12:20:33 发布

阅读量80

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/codingEskimo/p/6828846.html

本文介绍了一种高效算法来找到已排序数组中离目标值最近的K个数。该算法分为两部分：首先使用二分查找找到最接近目标值的数；然后通过双指针法找出其余K-1个最接近的数。文章提供了详细的实现代码，并讨论了一些特殊情况。

Important Node:

1) This problem has two parts:

a. find the closest number's to the target and return its index. This is standard binary search problem. O(logN)

b. find kth closest number. This is two pointers problem. O(k)

in this b part, be careful about the boundary. j should be less than nums.length instead of k, because rstIdx < k already control it.

2) Near to pay attension to coner case when k is equal to 0, just return an empty array instead of null

3) rst[rstIdx++] => rst[rstIdx] then rstIdx++

rst[++rstIdx] => rstIdx++ then rst[rstIdx(the one that already increased)]

public class Solution {
    /**
     * @param A an integer array
     * @param target an integer
     * @param k a non-negative integer
     * @return an integer array
     */
    public int[] kClosestNumbers(int[] A, int target, int k) {
        // Write your code here
        if (A == null || A.length == 0) {
            return null;
        }
        
        if (k == 0) {
            return new int[0];
        }
        
        //find the nearst number
        int nearIdx = findNearestIdx(A, target); 
        
        //find the first Kth closest number
        int[] rst = findKthNearest(A, k, target, nearIdx); 
        
        return rst;
    }
    
    private int findNearestIdx(int[] nums, int target) {
        int start = 0;
        int end = nums.length - 1;
        
        while (start + 1  < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        int startDis = Math.abs(nums[start] - target);
        int endDis = Math.abs(nums[end] - target);
        
        return startDis <= endDis ? start : end;
    }
    
    private int[] findKthNearest(int[] nums, int k, int target, int nearIdx) {
        int[] rst = new int[k];
        rst[0] = nums[nearIdx];
        int rstIdx = 1;
        int i = nearIdx - 1, j = nearIdx + 1;
        while (rstIdx < k && i >=0 && j < nums.length) {
            int disi = Math.abs(target - nums[i]);
            int disj = Math.abs(target - nums[j]);
            if (disi <= disj) {
                rst[rstIdx++] = nums[i--];
            } else {
                rst[rstIdx++] = nums[j++];
            }
        }
        
        while (rstIdx < k && i >= 0) {
            rst[rstIdx++] = nums[i--];
        }
        
        while (rstIdx < k && j < nums.length) {
            rst[rstIdx++] = nums[j++];
        }
        
        return rst;
    } 
    
}