[leetcode] Search in Rotated Sorted Array @ Python [Figure][自配插图说明]

浏览数：24 / 时间：2015年06月08日

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.

There two solutions:

Solution 1: How to hand the specific case involved point a,b and c?

Solution 2: How to hand the specific case involved point a, nd d?

Solution 1:

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be searched
    # @return an integer
    def search(self, A, target):
        return self.helper(A, target, 0, len(A)-1)
        
    def helper(self, A, target, low, high):
        if low > high: return -1
        mid = (low + high) / 2
        if target == A[mid]: return mid
        elif A[low] <= target < A[mid] or ( A[mid] <= A[high] and (target < A[mid] or target > A[high]) ):
            return self.helper(A, target,low, mid-1)
        else:
            return self.helper(A, target,mid+1, high)

Solution 2:

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be searched
    # @return an integer
    def search(self, A, target):
        low,high = 0,len(A)-1
        while low <= high:
            mid = (low+high+1)/2
            if A[mid]==target: return mid
            if A[low] <= A[mid]:
                if A[low] <= target < A[mid]: high=mid-1
                else: low=mid+1
            else:
                if A[mid] < target <= A[high]: low=mid+1
                else: high=mid-1
        return -1