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

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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。