Search in Rotated Sorted Array
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.
二分查找的推广，需要增加判断条件，为方便举例我们假设要找的数字为3，先用mid把数列分为两种情况，再讨论两种情况的子情况
mid = first+ ( end - first)/2;
if ( A[mid ] == target ) return mid;(找到终止)
if ( A[first] <= A[mid]) (这个时候还得考虑两种情况，1 2 3 4 5 6 7 8 9 和 5 6 7 8 9 1 2 3 4)
{
if( A[first] < target && target < =A[mid])
{ end=mid ; }
else
{ first= mid+1; }
}
else ./A[first] > A[mid]) (7 8 9 1 2 3 4 5 6 和 9 1 2 3 4 5 6 7 8)
{
if(A[mid] < target && target <=A[end-1])
{ first = mid+1 }
else
{ end = mid; }
}
class Solution {
public:
int search(int A[], int n, int target) {
int first=0;
int end=n;
while(first!=end)
{
int mid =first+ (end - first)/2;
if(A[mid] == target) return mid;
if(A[first] <= A[mid])
{
if(A[first] <= target && target<A[mid] )
{
end = mid;
}
else
{
first = mid+1;
}
}
else
{
if(A[mid] < target && target <= A[end-1] )
{
first = mid+1;
}
else
{
end = mid;
}
}
}
return -1;
}
};版权声明：本文为博主原创文章，未经博主允许不得转载。