leetcode33 Search in Rotated Sorted Array

"""
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:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
"""
"""
题目要求时间复杂度为O(log n)
先不管这个要求。AC混一下
得到解法一
"""
class Solution1:
    def search(self, nums, target):
        if target not in nums:
            return -1
        for i in range(len(nums)):
            if nums[i] == target:
                return i

"""
真正的解法是二分法
主要分两种情况：
    若nums[left]<=nums[mid]，这说明前半部分是排序的
    若nums[left]<=target<=nums[mid]，在前半部分搜索，反之在后半部分搜索
    若nums[left]>nums[mid]，这说明后半部分是排序的
    若nums[mid]<=target<=nums[right]，在后半部分搜索，反之在前半部分搜索
    一直搜索，直到left>=right
传送门：https://www.jianshu.com/p/7295418b260f
"""
class Solution2:
    def search(self, nums, target):
        n = len(nums)
        left, right = 0, n-1
        while left < right:
            mid = (left+right)//2
            if nums[mid] == target:
                return mid
            if nums[left] <= nums[mid]: #前办部分有序
                if nums[left] <= target <= nums[mid]:
                    right = mid
                else:
                    left = mid + 1
            else:#后半部分有序
                if nums[mid] <= target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid
        return left if nums and nums[left] == target else -1