题目描述：
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：
本题要求时间复杂度为 O(logn) ，自然应该想到二分查找。
首先，找到target的位置，然后分别从前从后找首尾位置即可，这是简单明了的解法。
注：在找到 target 后继续二分查找首尾出现的位置效率会更高一些。
代码实现：
class Solution {
public int[] searchRange(int[] nums, int target) {
int ret[] = {-1,-1};
int left = 0;
int right = nums.length-1;
int mid;
while(left<=right){
mid = (right+left)/2;
if(nums[mid]==target){
int l = mid;
int r = mid;
while(l>=0 && nums[l]==target){
l--;
}
while(r<=nums.length-1 && nums[r]==target){
r++;
}
ret[0] = l+1;
ret[1] = r-1;
return ret;
}else{
if(nums[mid]<target){
left = mid+1;
}else{
right = mid-1;
}
}
}
return ret;
}
}