二分查找是我们经常会遇到的算法，思路清晰，代码简洁。二分查找要求序列有序，且支持随机存取，一般情况下我们讨论的序列不存在相同元素，则二分查找可以很熟练的表示如下：
int binsearch(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
return mid;
}
}
return -1;
}
但是当我们对有序序列不做要求时，即可能出现相同的元素的情况下，二分查找就回出现一些扩展的问题，比如：
int searchLowerBound(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
res = mid;
right = mid-1;
}
}
return res;
} 对于（2），同理可得：int searchHigherBound(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
res = mid;
left = mid+1;
}
}
return res;
} 对于（3），使得key=key-1，则转换成问题（2），对于（4），使得key=key+1，则转换成问题（1）。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].
这个题就是要求出最小的value的下标和最大的value的下标，所以代码易得：
class Solution {
public:
int searchLowerBound(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
res = mid;
right = mid-1;
}
}
return res;
}
int searchHigherBound(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
res = mid;
left = mid+1;
}
}
return res;
}
vector<int> searchRange(int A[], int n, int target) {
vector<int> res;
res.push_back(searchLowerBound(A,n,target));
res.push_back(searchHigherBound(A,n,target));
return res;
}
};原文地址：http://blog.csdn.net/majing19921103/article/details/44808663
