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.
思路:颠倒后的序列从中间分开的话,至少有一端是增序的。
如果左边增序,且目标在这个范围内,则对左边进行二分查找,否则对右边进行。
class Solution {
public:
int search(int A[], int n, int target)
{
int left = 0;
int right = n-1;
while(left<=right)
{
int mid = (left + right) / 2;
if(target == A[mid])
return mid;
if(A[left] <= A[mid])
{
if (A[left] <= target && target < A[mid])
right = mid-1;
else
left = mid+1;
}
else
{
if (A[mid] < target && target <= A[right])
left = mid + 1;
else
right = mid-1;
}
}
return -1;
}
};
另外细节地方,如果初始化right = n,则wihile可写为(left!=right),内部 A[right] 改为 A[right-1] , mid-1 改为mid