【leetcode】33. Search in Rotated Sorted Array

题目如下：

解题思路：题目要求时间复杂度是O(log n)，而且数组也是有序的，那么可以考虑采用二分查找法。那么解题的关键就是找出转折点，找到了转折点后，把数组拆分成两段，再分别用二分查找，即可得到答案。

代码如下：

class Solution(object):def search(self, nums, target):""":type nums: List[int]:type target: int:rtype: int"""if len(nums) == 0:return -1split = 0for i in xrange(len(nums)-1):if nums[i] > nums[i+1]:split = i + 1breakimport  bisect#print nums[:split]res = bisect.bisect_left(nums[:split],target)#print resif res < len(nums[:split]) and nums[:split][res] == target:return resres = bisect.bisect_left(nums[split:], target)#print nums[split:]#print resif res < len(nums[split:]) and nums[split:][res] == target:return res + splitreturn -1

转载于:https://www.cnblogs.com/seyjs/p/8995983.html

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