Search for a Range

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int>res(2,-1);
if(nums.size()==0) return res;
int low=0,high=nums.size()-1,mid;
int mid1,high1,mid2,low2;
while(low<=high &&low>=0)
{
mid=(low+high)/2;
if(nums[mid]==target)
{
if(nums[low]==target)res[0]=low;
else
{
high1=mid;
while(low<high1)
{
mid1=(high1+low)/2;
if(nums[mid1]==target)
if(nums[mid1-1]!=target){res[0]=low=mid1;break;}
else high1=mid1-1;
else low=mid1+1;
}
res[0]=low;
}
if(nums[high]==target) res[1]=high;
else{
low2=mid;
while(low2<high)
{
mid2=(high+low2)/2;
if(nums[mid2]==target)
if(nums[mid2+1]!=target) {res[1]=low2=mid2;break;}
else low2=mid2+1;
else high=mid2-1;
}
res[1]=low2;
}
break;
}
else if(nums[mid]<target) low=mid+1;
else high=mid-1;
}
return res;
}
};

vector<int> searchRange(int A[], int n, int target) {
int i = 0, j = n - 1;
vector<int> ret(2, -1);
// Search for the left one
while (i < j)
{
int mid = (i + j) /2;
if (A[mid] < target) i = mid + 1;
else j = mid;
}
if (A[i]!=target) return ret;
else ret[0] = i;

// Search for the right one
j = n-1;  // We don't have to set i to 0 the second time.
while (i < j)
{
int mid = (i + j) /2 + 1;   // Make mid biased to the right
if (A[mid] > target) j = mid - 1;
else i = mid;               // So that this won't make the search range stuck.
}
ret[1] = j;
return ret;
}