难度: 中等(Medium)

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.

代码

class Solution:
    # @param {integer[]} nums
    # @param {integer} target
    # @return {integer}
    def search(self, nums, target):
        def binary_search(nums,l,r,target):
            if l > r:
                return -1
            m = l + (r-l)/2
            if nums[m] == target:
                return m
            elif nums[m] > target:
                return binary_search(nums,l,m-1,target)
            else:
                return binary_search(nums,m+1,r,target)

        i = 0
        while i < len(nums) -1:
            if nums[i+1] < nums[i]:
                break
            i += 1
        if nums[0] == target:
            return 0
        elif nums[0] > target:
            return binary_search(nums,i+1,len(nums)-1,target)
        else:
            return binary_search(nums,0,i,target)