problem：
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].
thinking：
（1）又是对二分搜索的变形，我的策略是当遇到 A[mid] = target时，以mid为中心往两边探测，搜索区间范围。
code：
class Solution {
public:
vector<int> searchRange(int A[], int n, int target) {
if(n==0)
{
res.push_back(-1);
res.push_back(-1);
return res;
}
if(n==1)
{
if(A[0]==target)
{
res.push_back(0);
res.push_back(0);
return res;
}
else
{
res.push_back(-1);
res.push_back(-1);
return res;
}
}//if
binary_search(A,0,n-1,target);
return res;
}
protected:
void binary_search(int A[], int left, int right, int target)
{
if(left>right)
{
res.push_back(-1);
res.push_back(-1);
return;
}
int mid = (left+right)/2;
if(A[mid]==target)
{
int i=mid,j=mid;
while(i>=left && i<=right && A[i]==target)
i--;
res.push_back(++i);
while(j>=left && j<=right &&A[j]==target)
j++;
res.push_back(--j);
return;
}
else if(A[mid]>target)
binary_search(A,left,mid-1,target);
else
binary_search(A,mid+1,right,target);
}
private:
vector<int> res;
};leetcode 题解 || Search for a Range 问题
原文：http://blog.csdn.net/hustyangju/article/details/44617583