leetcode 【 Search in Rotated Sorted Array II 】python 实现

题目：

与上一道题几乎相同；不同之处在于array中允许有重复元素；但题目要求也简单了，只要返回true or false

http://www.cnblogs.com/xbf9xbf/p/4254590.html

 

代码：oj测试通过 Runtime: 73 ms

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

 

思路：

用了一个trick Python数组去重的办法A=list(set(A))

这样A数组中就没有重复的元素了，可以直接用之前一题的代码。

这样的trick应该不是题目的本意，这道二分查找题目比较经典，应该吃透。