leetcode 【 Search in Rotated Sorted Array 】python 实现

题目：

Suppose a sorted array 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

 

代码：oj测试通过 Runtime: 53 ms

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be searched
    # @return an integer
    def search(self, A, target):
        # none case & zero case
        if A is None or len(A)==0 :
            return -1
        # binary search
        start = 0
        end = len(A)-1
        while start<=end :
            # one element left case
            if start == end :
                if A[start]==target :
                    return start
                else:
                    return -1
            # two elements left case
            if start+1 == end :
                if A[start]==target :
                    return start 
                elif A[end]==target :
                    return end
                else:
                    return -1
            # equal or more than three elements case
            mid = (start+end)/2
            if A[mid]==target :
                return mid
            elif A[mid]>target:
                if A[start]>A[mid] and A[end]<A[mid]:
                    start = mid+1
                elif A[start]<A[mid] and A[end]<A[mid]:
                    if A[end]>=target:
                        start = mid+1
                    else:
                        end = mid-1
                elif A[start]>A[mid] and A[end]>A[mid]:
                    end = mid-1
                else:
                    end = mid-1
            else:
                if A[start]>A[mid] and A[end]<A[mid]:
                    end = mid-1
                elif A[start]<A[mid] and A[end]<A[mid]:
                    start = mid+1
                elif A[start]>A[mid] and A[end]>A[mid]:
                    if A[end]>=target :
                        start = mid+1
                    else:
                        end = mid-1
                else:
                    start = mid+1
        return -1

思路：

这个就是binary search的思路。

个人没想出来什么好的方法，硬着头皮硬写了一个暴力解决方法。

传统的binary search只需要判断A[mid]与target的大小就可以了；但这道题是rotated array，光判断A[mid]是不够的。

还需要判断A[start] A[end]与A[mid]的大小才能判断，target可能落在[start,mid]区间还是[mid,end]区间。

自己的代码实在有些繁琐丑陋，估计有些if else条件可以合并，后续会改进。