081 Search in Rotated Sorted Array II
在 033 Search in Rotated Sorted Array 稍作改进， 但是最坏情况下 复杂度为O(n)
class Solution:# @param A, a list of integers# @param target, an integer to be searched# @return an integerdef search(self, A, target):if len(A) == 0:return Falsereturn self.solve(A, target, 0, len(A)- 1)def solve(self, a, target, start, end):if start > end:return Falsehalf = int((end-start)/2) + startif a[start] == target:return Trueif a[end] == target:return Trueif a[half] == target:return Trueif a[start] < a[half]:if target > a[half] or target < a[start]:return self.solve(a, target, half+1, end-1)else:return self.solve(a, target, start+1, half-1)elif a[start] > a[half]:if target < a[half] or target > a[start]:return self.solve(a, target, start+1, half-1)else:return self.solve(a, target, half+1, end-1)else:return self.solve(a, target, start+1, half-1) or self.solve(a, target, half+1, end-1)
转载于:https://www.cnblogs.com/dapanshe/p/4629870.html
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