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
1) 这个题目思路就是先画图, 分区间讨论,注意是左移还是右移,最好画图判断下。
2)可以利用[LeetCode] 153. Find Minimum in Rotated Sorted Array_Medium tag: Binary Search 的思路,通过lg n 找到转折点,然后分别在两段里面找。
T: O(lgn)
Code
1)
class Solution: def search(self, nums, target): if not nums: return -1 S, E, l, r = nums[0], nums[-1], 0, len(nums) - 1 while l + 1 < r: mid = l + (r - l)//2 if target == nums[mid]: return mid if nums[mid] >= S: if S <= target <= nums[mid]: r = mid else: l = mid else: if nums[mid] <= target <= E: l = mid else: r = mid if nums[l] == target: return l if nums[r] == target: return r return -1
原文地址:https://www.cnblogs.com/Johnsonxiong/p/10715364.html
时间: 2024-11-09 01:45:15