[leetcode] Search for a Range

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]

.

分析：二分查找

class Solution
{
public:
void Searching(vector<int> &ret, int A[], int s, int e, int target)
{
if(s > e)
{
ret.push_back(-1);
ret.push_back(-1);
return;
}

int mid = (s+e)/2;
if(A[mid] < target)
Searching(ret, A, mid+1, e, target);
else if(A[mid] > target)
Searching(ret, A, s, mid-1, target);
else
{
int i = mid - 1;
while(i >= s && A[i] == A[mid]) i--;
ret.push_back(i + 1);

i = mid + 1;
while(i <= e && A[i] == A[mid]) i++;
ret.push_back(i - 1);
}
}
vector<int> searchRange(int A[], int n, int target)
{
vector<int> ret;
if(target < A[0] || target > A[n-1])
{
ret.push_back(-1);
ret.push_back(-1);
}
else
Searching(ret, A, 0, n-1, target);

return ret;
}
};