java算法之Search for a Range

问题：Search for a Range

Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

思路：因为题目要求用 O(log n)时间复杂度，所以我们考虑用二分查找方法，我们只需要找到一个target的位置，

然后指定两个指针，分别从target位置依次向前向后遍历，找到所有的target值就好。。题目不难

代码：

import java.util.Arrays;public class SearchOfRange {public static void main(String[] args) {int[] a=new int[] {8,8,8,8,8,10};int target=8;int []result=searchRange(a, target);System.out.println(Arrays.toString(result));}public static int[] searchRange(int []a, int target) {int end=a.length;int start=0;int mid=0;while(start<end) { mid=(end+start)/2;if(a[mid]<target) {end=mid-1;}if(a[mid]>target) {start=mid+1;}else break;}int []result=new int[2];if(a[mid]==target) {int rangeStart=mid;int rangeEnd=mid;//这里rangStart不能等于0,如果等于0的话，rangeStart-1则为-1while(rangeStart>0&&a[rangeStart-1]==target) rangeStart--;while(rangeEnd<a.length-1&&a[rangeEnd+1]==target) rangeEnd++;result[0]=rangeStart;result[1]=rangeEnd;}else {result[0]=-1;result[1]=-1;}return result;}}