Leetcode** 81. Search in Rotated Sorted Array II

Description: There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).

Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].

Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.

Link: 81. Search in Rotated Sorted Array II

Examples:

Example 1:
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example 2:
Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Example 3:
Input: nums = [1,0,1,1,1], target = 0
Output: true

思路: 这个题目是33的加强版，添加了重复的数字，但只要返回True, False. 如果按照33的源代码就会过不了example3，因为l == mid, mid == r,就没办法进入看到0.返回False。症结就在于数组断开的地方前后是重复的数字，一部分留在开头，另一部分留在了末尾，因为只要返回是否存在，所以l从不同与r的数开始，再开始正常的搜索。

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: bool
        """
        if not nums: return False
        n = len(nums)
        l, r = 0, n-1
        while l < r and nums[l] == nums[r]:
            l += 1
        while l <= r:
            mid = int((l+r)/2)
            if target == nums[mid]:
                return True
            if nums[l] <= nums[mid]:
                if target >= nums[l] and target <= nums[mid]:
                    r = mid - 1
                else:
                    l = mid + 1 
            elif nums[mid] <= nums[r]:
                if target >= nums[mid] and target <= nums[r]:
                    l = mid + 1
                else:
                    r = mid - 1
        return False

在这篇题解博客中，将找到l不同于r放在了while里面:

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: bool
        """
        if not nums: return False
        n = len(nums)
        l, r = 0, n-1
        while l <= r:
            while l < r and nums[l] == nums[r]:
                l += 1
            mid = int((l+r)/2)
            if target == nums[mid]:
                return True
            if nums[l] <= nums[mid]:
                if target >= nums[l] and target <= nums[mid]:
                    r = mid - 1
                else:
                    l = mid + 1 
            elif nums[mid] <= nums[r]:
                if target >= nums[mid] and target <= nums[r]:
                    l = mid + 1
                else:
                    r = mid - 1
        return False

日期: 2021-04-11 96岁的励志奶奶尚且如此辛苦