Search in Rotated Sorted Array

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.

Example

For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

Solution

S: 上道题最重要的是找到，Target的值，而这道题就是要确认中点的位置，在左上象限 或者是右下象限，然后根据Target的落入点进行排出，巧妙之处就是逐步排出，先把能排除的都排除掉，而左上mid 后面的可以慢慢的推出。

class Solution:
    """
    @param A : a list of integers
    @param target : an integer to be searched
    @return : an integer
    """
    def search(self, num, target):
        # write your code here
        if len(num) == 0:
            return -1

        start, end = 0, len(num) - 1
        while start + 1 < end:
            mid = ( start + end ) / 2
            if num[mid] == target:
                return mid
            if num[mid] > num[start]:
                if target >= num[start] and target <= num[mid]:
                    end = mid
                else:
                    start = mid
            else:
                if target >= num[mid] and target <= num[end]:
                    start = mid
                else:
                    end = mid
        if num[start] == target:
            return start
        if num[end] == target:
            return end
        return -1