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 }
leetcode 之Search in Rotated Sorted Array（三）
原文：http://www.cnblogs.com/573177885qq/p/5451171.html