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本题的思路是二分查找，关键是怎么用二分查找。通过middle值和数组尾部的值比较，可以确定start-Middle和middle-end，这两部分那一部分是有序的，有序的数组是可以用二分查找的。
class Solution {
public:
int search(int A[], int n, int target) {
if(n == 0)
return 0;
int middle,end,start;
start = 0;
end = n-1;
middle = (start + end)/2;
int result;
while(start <= end)
{
if(A[middle] < A[end])
{
result = findbinarysearch(A,middle,end,target);
if(result !=-1)
return result;
end = middle -1;
middle = (start+ end)/2;
}
else if(A[middle] > A[end])
{
result = findbinarysearch(A,start,middle,target);
if(result !=-1)
return result;
start = middle +1 ;
middle = (start+ end)/2;
}
else
{
if(A[middle] == target)
return middle;
else
return -1;
}
}
return -1;
}
int findbinarysearch(int A[],int start, int end, int target)
{
int middle = (start + end)/2;
while(start <= end)
{
if(A[middle] < target)
{
start = middle +1;
middle = (start + end)/2;
}
else if(A[middle] > target)
{
end = middle -1;
middle = (start + end)/2;
}
else
{
return middle;
}
}
return -1;
}
};
Search in Rotated Sorted Array
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原文地址：http://www.cnblogs.com/xgcode/p/4265550.html
