Q: Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
A: 递归+二分查找。
int bisearch(int A[],int begin,int end,int target)
{
if(begin>end)
return -1;
int middle = begin+(end-begin)/2;
if(A[middle] == target)
return middle;
if(A[begin]>A[end])
{
int pos1 = bisearch(A,begin,middle-1,target);
int pos2 = bisearch(A,middle+1,end,target);
return (pos1!=-1?pos1:pos2);
}else
{
//这里不能写成return bisearch(A,begin,end,target);---这样相当于搜索的范围永远不会减少。。死循环
if(A[middle]>target)
return bisearch(A,begin,middle-1,target);
else
return bisearch(A,middle+1,end,target);
}
}
int search(int A[], int n, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return bisearch(A,0,n-1,target);
}
原文链接: https://www.cnblogs.com/summer-zhou/p/3141213.html
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