Leetcode126: 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]

.

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int n = nums.size();
int l = 0;
int r = n-1;
int res = -1;
vector<int> ret;
while(l<=r)
{
int mid = (l+r)/2;
if(nums[mid] == target)
{
res = mid;
break;
}
else if(nums[mid] < target)
l = mid+1;
else
r = mid-1;
}
if(res == -1)
{
ret.push_back(-1);
ret.push_back(-1);
return ret;
}
int ls=res;
int rs=res;
while(ls>=0 && nums[ls] == nums[res])
{
ls--;
}
ls++;
while(rs<=n-1 && nums[rs] == nums[res])
{
rs++;
}
rs--;
ret.push_back(ls);
ret.push_back(rs);
return ret;
}
};