Search in Rotated Sorted Array-优快云博客

本文链接：https://blog.youkuaiyun.com/youqika/article/details/41292521

本文介绍了一种在已知没有重复元素的旋转排序数组中搜索特定目标值的方法。通过定义一个解决方案类并实现search方法，该算法能够高效地找到目标值在数组中的位置，如果未找到则返回-1。此外，还提供了查找最小元素位置的辅助方法，以及针对不同子数组进行二分搜索的实现。

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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.

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be searched
    # @return an integer
    def search(self, A, target):
        if 0 == len(A):
            return -1
        minx = self.__findMin(A)
        if target < A[minx] or target > A[minx - 1]:
            return -1
        rslt = 0
        if target >= A[minx] and target <= A[-1]:
            rslt = self.binsearch(A[minx:], target)
            if -1 == rslt:
                return -1
            rslt += minx
        else:
            rslt = self.binsearch(A[:minx], target)
        return rslt


    def binsearch(self, array, target):
        if 0 == len(array):
            return -1

        mid = len(array) / 2
        if target == array[mid]:
            return mid
        elif len(array) > 1:
            if target < array[mid]:
                return self.binsearch(array[:mid], target)
            else:
                rslt = self.binsearch(array[mid:], target)
                if -1 == rslt:
                    return -1
                return mid + rslt
        else:
            return -1

    def __findMin(self, num):
        if num[-1] >= num[0]:
            return 0
        mid = len(num) / 2
        if num[mid] < num[0]:
            return 1 + self.__findMin(num[1:mid+1])
        else:
            return mid + 1 + self.__findMin(num[mid+1:])

s = Solution()
arr = [5,1,3]
#print s.search(arr, 2)
print s.binsearch(arr, 2)