33. Search in Rotated Sorted Array(Python3)

33. Search in Rotated Sorted Array(Python3)

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.

解题方案

思路：

• 二分法查找，时间复杂度到 log(n) l o g ( n )
• 这里边界的处理很容易出错，在两个示例上一直出错，一个是 [1,3],1 [ 1 , 3 ] , 1 ，还有一个 [3,1],1 [ 3 , 1 ] , 1

代码：

class Solution:
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        front = 0;
        end = len(nums)-1
        while front <= end:
            m = (end + front)//2
            if target == nums[m]:
                return m
            if nums[front] <= nums[m]:
                if target < nums[m] and target >= nums[front]:
                    end = m-1
                else:
                    front = m+1
            else:
                if target <= nums[end] and target > nums[m]:
                    front = m+1
                else:
                    end = m-1
        return -1