题目

题目链接：
https://www.nowcoder.com/practice/b4d7edc45759453e9bc8ab71f0888e0f

知识点

	二分查找；找到第一个大于等于x的数的位置idx;然后从idx开始往两边扩展

Java代码

import java.util.*;public class Solution {/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param nums int整型ArrayList* @param k int整型* @param x int整型* @return int整型ArrayList*/public ArrayList<Integer> closestElement (ArrayList<Integer> nums, int k,int x) {//二分+双指针int n = nums.size();int right = firstGt(nums, x);int left = right - 1;ArrayList<Integer> ans = new ArrayList<>();LinkedList<Integer> ll = new LinkedList<>();if (right == 0) {for (int i = 0; i < k ; i++) {ll.add(nums.get(i));}} else if (right == n - 1) {for (int i = n - 1 - k; i < n ; i++) {ll.add(nums.get(i));}} else {while (left >= 0 || right < n) {int diffleft = -1;int diffright = -1;if (left >= 0) {diffleft = x - nums.get(left);}if (right < n) {diffright = nums.get(right) - x;}if (diffleft != -1 && diffright != -1) {if (diffleft <= diffright) {ll.addFirst(nums.get(left--));} else {ll.addLast(nums.get(right++));}} else if (diffleft != -1) {ll.addFirst(nums.get(left--));} else if (diffright != -1) {ll.addLast(nums.get(right++));}if (ll.size() == k)break;}}return new ArrayList<>(ll);}//找到大于等于x的下标位置public int firstGt(ArrayList<Integer> nums, int x) {int n = nums.size();int left = 0;int right = n;while (left < right) {int m = left + (right - left) / 2;if (nums.get(m) > x) {right = m - 1;} else if (nums.get(m) < x) {left = m + 1;} else {return m;}}return left;}
}

Go代码

package main//import "fmt"/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param nums int整型一维数组* @param k int整型* @param x int整型* @return int整型一维数组*/
func closestElement(nums []int, k int, x int) []int {//二分+双指针n := len(nums)right := firstGt(nums, x)arrleft := []int{}arrright := []int{}ans := []int{}if right == 0 {for i := 0; i < k; i++ {ans = append(ans, nums[i])}} else if right == n-1 {for i := n - 1 - k; i < n; i++ {ans = append(ans, nums[i])}} else {left := right - 1cnt := 0for left >= 0 || right < n {diffleft := -1diffright := -1if left >= 0 {diffleft = x - nums[left]}if right < n {diffright = nums[right] - x}if diffleft != -1 && diffright != -1 {if diffleft <= diffright {arrleft = append(arrleft, nums[left])left--} else {arrright = append(arrright, nums[right])right++}} else if diffleft != -1 {arrleft = append(arrleft, nums[left])left--} else if diffright != -1 {arrright = append(arrright, nums[right])right++}cnt++if cnt == k {for i := len(arrleft) - 1; i >= 0; i-- {ans = append(ans, arrleft[i])}for i := 0; i < len(arrright); i++ {ans = append(ans, arrright[i])}break}}}return ans
}//找到大等于x的位置
func firstGt(nums []int, x int) int {n := len(nums)left := 0right := n - 1for left < right {m := left + (right-left)/2if nums[m] > x {right = m - 1} else if nums[m] < x {left = m + 1} else {return m}}return left
}

PHP代码

<?php/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可** * @param nums int整型一维数组 * @param k int整型 * @param x int整型 * @return int整型一维数组*/
function closestElement( $nums ,  $k ,  $x )
{// 二分+双指针$n = count($nums);$right = firstGt($nums,$x);$ans = [];$arrleft=[];$arrright =[];if($right ==0 ){for($i=0;$i<$k;$i++){$ans[$i] = $nums[$i];}}else if($right ==$n-1){for($i=$n-1-$k;$i>=0;$i++){$ans[count($ans)] = $nums[$i];}}else {$left = $right-1;$cnt =0;while ($left>=0 || $right < $n){$diffleft=-1;$diffright =-1;if($left>=0) {$diffleft = $x-$nums[$left];}if($right<$n){$diffright = $nums[$right]-$x;}if($diffleft!=-1 && $diffright!=-1){if($diffleft<=$diffright){$arrleft[count($arrleft)] = $nums[$left--];}else{$arrright[count($arrright)] = $nums[$right++];}}else if($diffleft!=-1){$arrleft[count($arrleft)] = $nums[$left--];}else if($diffright!=-1){$arrright[count($arrright)] = $nums[$right++];}$cnt++;if($cnt==$k){for($i=count($arrleft)-1;$i>=0;$i--){$ans[count($ans)] = $arrleft[$i];}for($i=0;$i<count($arrright);$i++){$ans[count($ans)] = $arrright[$i];}break;}}}return $ans;
}//找到大于等于x的位置
function firstGt($nums,$x){$n = count($nums);$left =0;$right = $n;while ($left<$right){$m = $left+(($right-$left)>> 1);if($nums[$m] > $x){$right = $m-1;}else if($nums[$m] < $x){$left = $m+1;}else{return $m;}}return $left;
}