题目：
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.
Example
For [4, 5, 1, 2, 3] and target=1, return 2.
For [4, 5, 1, 2, 3] and target=0, return -1.
题解：
class Solution { public: /** * @param nums: a rotated sorted array * @return: the minimum number in the array */ int search(vector<int> &A, int target) { if (A.empty()) { return -1; } int start = 0, end = A.size() - 1; while (start + 1 < end) { int mid = start + (end - start) / 2; if (A[mid] == target) { return mid; } if (A[start] < A[mid]) { if (A[start] <= target && target <= A[mid]) { end = mid; } else { start = mid; } } else { if (A[mid] <= target && target <= A[end]) { start = mid; } else { end = mid; } } } if (A[start] == target) { return start; } else if (A[end] == target){ return end; } return -1; } };
【Lintcode】062.Search in Rotated Sorted Array
原文：http://www.cnblogs.com/Atanisi/p/6822611.html