【92】【H】【leetcode】 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.

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关键就是，找到最小值的坐标之后，用pivot+mid 取模作为真实的mid

小技巧，如果while后面的a<b，那就a = mid +1,

如果while后面是a<= b, 那就a = mid+1,b = mid - 1

class Solution(object):    def search(self, nums, target):        a = 0        b = len(nums) - 1        while a<b:            mid = (a + b) / 2            if nums[mid] < nums[b]:                b = mid            else :                a = mid + 1        #print nums[a]        pilot = a        #print pilot         a = 0        l = len(nums)        b = l - 1        while a <= b:            mid = (a+b)/2            realmid = (mid + pilot) % (l-0)           # print a,b,mid,realmid            if nums[realmid] < target:                a = mid + 1            elif nums[realmid] > target:                b = mid - 1            else:                return realmid        return -1        """        :type nums: List[int]        :type target: int        :rtype: int        """