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].
vector<int> searchRange(int A[], int n, int target){
int l = distance(A, lower_bound(A, A + n, target));
int u = distance(A, upper_bound(A, A + n, target));
vector<int> res;
if(A[l] != target){
res.push_back(-1);
res.push_back(-1);
}else{
res.push_back(l);
res.push_back(u - 1);
}
return res;
}
leetcode 二分查找 Search for a Range
原文：http://blog.csdn.net/zhengsenlie/article/details/39969821