描述
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.
二分法查找适用于排好序的,所以这题的关键是如何确定某部分的顺序是怎样的。
1 int searchRotateSA(int A[], int n,int target) 2 { 3 int first = 0, last = n; 4 while (first!=last) 5 { 6 int mid = first + (last - first) / 2; 7 if (A[mid] = target) 8 { 9 return mid; 10 } 11 else if (A[first]<=A[mid])//判断大小顺序 12 { 13 if (A[first] <= target&&target <= A[mid]) 14 last = mid; 15 else 16 first = mid + 1; 17 } 18 else 19 { 20 if (A[first] >= target&& A[mid] <= target) 21 last = mid; 22 else 23 first = mid + 1; 24 25 } 26 } 27 28 return -1; 29 }
时间: 2024-10-12 18:14:53