Leetcode -- 33. Search in Rotated Sorted Array

题目：

Suppose an array sorted in ascending order 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.

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思路：由于数组是有递增有序数组旋转得到的，我们需要找到有序数组的旋转轴，然后对旋转后的数组分段利用二分查找。

1. 利用二分查找，找到旋转轴；
2. 对两段递增数组利用二分查找。

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C++代码如下：

class Solution {public:    int search(vector<int>& nums, int target) {        if(nums.empty())            return -1;        int l = 0, r = nums.size()-1;        int i;        while (l <= r)        {            i = (l + r) / 2;            if (nums[i] > nums[i + 1])                break;            else if (nums[i] >= nums[0])                l = i + 1;            else                r = i - 1;        }        if (target >= nums[0] && target <= nums[i])            return binarySearch(nums, 0, i, target);        if (target >= nums[i + 1] && target <= nums[nums.size() - 1])            return binarySearch(nums, i + 1, nums.size() - 1, target);        return -1;    }private:    int binarySearch(vector<int>& nums, int l, int r, int target)    {        int mid;        while (l <= r)        {            mid = (l + r) / 2;            if (nums[mid] == target)                return mid;            else if (nums[mid] < target)                l = mid + 1;            else                r = mid - 1;        }        return -1;    }};