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.

Solution:

public class Solution {
    public int search(int[] nums, int target) {
        int index=0,len=nums.length;
        while(index<len-1 && nums[index]<=nums[index+1])index++;
        if(target>=nums[0] && target<=nums[index]){
            return find(nums,0,index,target);
        }else{
            return find(nums,index+1,len-1,target);
        }
    }
    private int find(int[] nums,int start,int end,int target){
        if(start>end)return -1;
        int l=start,r=end,mid;
        while(l<=r){
            mid=(l+r)/2;
            if(nums[mid]==target)return mid;
            else if(target<nums[mid])r=mid-1;
            else l=mid+1;
        }
        return -1;
    }
}

C语言源代码（3ms）：

int find(int* nums,int start,int end,int target){
    int l=start,r=end,mid;
    if(start>end)return -1;
    while(l<=r){
        mid=(l+r)>>1;
        if(nums[mid]==target)return mid;
        else if(nums[mid]>target)r=mid-1;
        else l=mid+1;
    }
    return -1;
}
int search(int* nums, int numsSize, int target) {
    int index=0;
    while(index<numsSize-1 && nums[index]<=nums[index+1])index++;
    if(target >= nums[0] && target<=nums[index]){
        return find(nums,0,index,target);
    }else{
        return find(nums,index+1,numsSize-1,target);
    }
}

C++源代码（6ms）：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int index=0,len=nums.size();
        while(index<len-1 && nums[index]<=nums[index+1])index++;
        if(target>=nums[0] && target<=nums[index]){
            return find(nums,0,index,target);
        }else{
            return find(nums,index+1,len-1,target);
        }
    }
private:
    int find(vector<int>& nums,int start,int end,int target){
        if(start>end)return -1;
        int l=start,r=end,mid;
        while(l<=r){
            mid=(l+r)>>1;
            if(nums[mid]==target)return mid;
            else if(nums[mid]>target)r=mid-1;
            else l=mid+1;
        }
        return -1;
    }
};

Python源代码（64ms）：

class Solution:
    # @param {integer[]} nums
    # @param {integer} target
    # @return {integer}
    def search(self, nums, target):
        index=0;length=len(nums)
        while index<length-1 and nums[index]<nums[index+1]:index+=1
        if target>=nums[0] and target<=nums[index]:return self.find(nums,0,index,target)
        else:return self.find(nums,index+1,length-1,target)
    def find(self,nums,start,end,target):
        if start>end:return -1;
        l=start;r=end
        while l<=r:
            mid=(l+r)/2
            if nums[mid]==target:return mid
            elif target>nums[mid]:l=mid+1
            else:r=mid-1
        return -1