81. Search in Rotated Sorted Array II

第一种做法

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: bool
        """
        if len(nums) == 0:
            return False
        if len(nums) == 1 and target != nums[0]:
            return False
        if len(nums) == 1 and target == nums[0]:
            return True
        q = 0
        for i in range(len(nums)):
            if i > 0 and nums[i] < nums[i-1]:
                q = i
                break
        if q == 0:
            return self.half(nums, target)
        elif target >= nums[0]:
            return self.half(nums[:q], target)
        else:
            return self.half(nums[q:], target)
    def half(self, nums, target):
        l = 0
        r = len(nums) - 1
        while l <= r:
            m = (l + r)/2
            if nums[m] == target:
                return True
            elif nums[m] > target:
                r = m -1
            else:
                l = m + 1
        return False

这一题和Ⅰ一样，因此第二种方法

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: bool
        """
        if len(nums) == 0:
            return False
        if len(nums) == 1 and target != nums[0]:
            return False
        if len(nums) == 1 and target == nums[0]:
            return True
        l = 0
        r = len(nums) - 1
        while l <= r:
            m = (l + r) / 2
            if nums[m] == target:
                return True
            if nums[m] > nums[l]:
                if nums[l] <= target <= nums[m]:
                    r = m - 1
                else:
                    l = m + 1
            elif nums[m] < nums[l]:
                if nums[m] <= target <= nums[r]:
                    l = m + 1
                else:
                    r = m - 1
            else:
                l += 1
        return False