LeetCode 33. Search in Rotated Sorted Array

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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.

Your algorithm’s runtime complexity must be in the order of O(log n).

Example 1:

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Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

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Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

思路

二分查找。
不过此题是Rotated Sorted Array，所以不能通过mid与左右指针指向的值的比较来移动指针。而是通过判断哪一部分是有序的，target是否在这个有序的部分来实现。

代码

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class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        if not nums:
            return -1
        left, right = 0, len(nums)-1
        while left <= right:
            m = (left+right)/2
            if target == nums[m]:
                return m
            elif nums[m] < nums[right]:
                if target > nums[m] and target <= nums[right]:
                    left = m + 1
                else:
                    right = m - 1
            else:
                if target >= nums[left] and target < nums[m]:
                    right = m - 1
                else:
                    left = m + 1
        return -1