Search for a Range

Given a sorted array of integers, 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].

思路就是扫一遍，如果出现等于target的地方开始记录，用while循环记录有多少个重复的target元素，while循环结束后返回最后一个等于target值的元素的下标

代码：

public class Solution {    public int[] searchRange(int[] nums, int target) {        //sorted arrayint res[] = new int[2];res[0]=-1;res[1]=-1;for(int i=0;i<nums.length;i++){if(nums[i]==target){//进入循环res[0]=i;int temp = i+1;while(temp<nums.length){if(nums[temp]==target)temp++;else break;}res[1]=temp-1;return res;}}return res;    }}