LeetCode:33. Search in Rotated Sorted Array

LeetCode:33. 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

就是有一个升序排列的数组，但是它从某个位置截成两半并交换了两部分的位置。要从它这里面找到目标值，存在则返回该值所处位置，否则返回-1。同事要求时间负责度不超过O(log n)。

思路：二分查找

由于限制了时间，所以不能遍历查找。先考虑一下二分查找。由于数组不是原来的升序，但是两部分内部还是有序的。原始二分查找的方法是查找时只继续查找符合条件的一部分，而我们这里，两部分都要执行二分查找。

Python 代码实现

class Solution:
    
    def searchPart(self, nums: List[int], left,right,target: int) -> int:
        l = len(nums)
      
        ans = -1
        if l == 0:
            return ans
        if left == right:
            if nums[left] == target :
                return left
            else:
                return ans
        mid = (int)((left+right)/2)
        if nums[mid] == target:
            ans = mid
        else:
            if left < mid :
                l_ans = self.searchPart(nums,left,mid-1,target)
                if l_ans != -1:
                    ans = l_ans
            if mid < right:
                r_ans = self.searchPart(nums,mid+1,right,target)
                if r_ans != -1:
                    ans = r_ans
     
        return ans
      
    def search(self, nums: List[int], target: int) -> int:
        return self.searchPart(nums, 0, len(nums)-1, target)

这里定义了一个 searchPart() 方法，查找时同时对mid两边执行。

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THE END.