Leetcode 33. Search in Rotated Sorted Array

https://leetcode.com/problems/search-in-rotated-sorted-array/

Medium

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

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• 二分查找。理解二分查找的原理，若在单调区间内，则正常二分缩小范围，否则针对边界进行变种的修改。通过坐标图能更容易理解算法。
• 需要注意code里两次用到<=的地方
  • while left <= right，保证了如果循环结束，结果必然是left > right，i.e. 没有找到target。
  • if nums[left] <= nums[middle]:，因为middle = (left + right) // 2是向下取整，所以需要用<=，e.g. [2, 0], 0
• https://leetcode.com/problems/search-in-rotated-sorted-array/discuss/14436/Revised-Binary-Search
• 6. Expressions — Python 3.7.4 documentation
  • https://docs.python.org/3/reference/expressions.html#comparisons
  • Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false).

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if not nums:
            return -1
        
        left, right = 0, len(nums) - 1
        
        # pay attention to <= 
        while left <= right:
            middle = (left + right) // 2
            
            if nums[middle] == target:
                return middle
            
            # pay attention to <=
            if nums[left] <= nums[middle]:
                if nums[left] <= target < nums[middle]:
                    right = middle - 1
                else:
                    left = middle + 1
            else:
                if nums[middle] < target <= nums[right]:
                    left = middle + 1
                else:
                    right = middle - 1
        
        return -1