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.

 

先用二分法找到pivot。判断target应该是在前一段还是后一段。然后在相应的那一段做二分法搜索。注意L39, 如果l - r ==1 这说明r就是pivot了，不能一直搜到 l == r.

 1 class Solution(object):
 2     def search(self, nums, target):
 3         """
 4         :type nums: List[int]
 5         :type target: int
 6         :rtype: int
 7         """
 8         pivot = self.findP(nums)
 9         if nums[pivot] <= target <= nums[-1]:
10             ans = self.BS(nums[pivot:], target)
11             if ans != -1:
12                 return ans + pivot
13             else:
14                 return -1
15         elif nums[0] <= target <= nums[pivot-1]:
16             return self.BS(nums[:pivot], target)
17         else:
18             return -1
19     
20     
21     def BS(self, nums, target):
22         n = len(nums)
23         l, r = 0, n-1
24         while l <= r:
25             mid = (l + r)//2
26             if nums[mid] == target:
27                 return mid
28             elif nums[mid] < target:
29                 l = mid + 1
30             else:
31                 r = mid - 1
32         return -1
33         
34     def findP(self, nums):
35         n = len(nums)
36         if nums[0] <= nums[n-1]:
37             return 0
38         l, r = 0, n -1
39         while l < r and r - l > 1:
40             mid = (l+r)//2
41             if nums[l] > nums[mid]:
42                 r = mid 
43             elif nums[mid] > nums[r]:
44                 l = mid 
45         return r 
46

 

posted @ 2017-01-09 13:10  lettuan  阅读(312)  评论(0)  编辑  收藏  举报