Medium 34题 Search for a Range

Question:

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]

.

Solution:we have a different loop condition here and we also apply binary search here to find the range.
public class Solution {
public int[] searchRange(int[] nums, int target) {
int low=0;
int high=nums.length-1;
int[] res=new int[2];
while(nums[low]<nums[high])
{
int mid=(high-low)/2+low;
if(target<nums[mid])
high=mid-1;
else if(target>nums[mid])
low=mid+1;
else
{
if(nums[low]==nums[mid])
high--;
else
low++;
}
}
if(nums[low]==nums[high]&&nums[low]==target)
{
res[0]=low;
res[1]=high;
}
else
{
res[0]=-1;
res[1]=-1;
}
return res;
}
}