81. Search in Rotated Sorted Array II

Search in Rotated Sorted Array II - LeetCode

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__.__ You must decrease the overall operation steps as much as possible.

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

  • code
class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        l, r = 0, len(nums) - 1
        
        while l <= r:
            while l < r and nums[l] == nums[l + 1]:
                l += 1
            while l < r and nums[r] == nums[r - 1]:
                r -= 1
                
            mid = l + (r - l) // 2
            if target == nums[mid]: return True
            elif nums[l] <= nums[mid]: # left part sorted, case for == [1,3]
                if nums[l] <= target < nums[mid]:
                    r = mid - 1
                else:
                    l = mid + 1
            else: # right part sorted
                if nums[mid] < target <= nums[r]:
                    l = mid + 1
                else:
                    r = mid - 1
        return False
  • code
class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        l, r = 0, len(nums) - 1
        
        while l <= r:
            mid = l + (r - l) // 2
            if target == nums[mid]: return True
            if nums[mid] > nums[l]: # first half sorted
                if nums[l] <= target < nums[mid]: # and target in
                    r = mid - 1
                else:
                    l = mid + 1
            elif nums[mid] < nums[r]: # second half sorted
                if nums[mid] < target <= nums[r]:
                    l = mid + 1
                else:
                    r = mid - 1
            else: # nums[end] <= nums[mid] <= nums[start]
                return target in set(nums)
                    
        return False
  • code
class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        start, end = 0, len(nums) - 1
        while start <= end:
            mid = start + (end - start) // 2
            if nums[mid] == target: return True
            if nums[start] < nums[mid]: # first half sorted
                if nums[start] <= target < nums[mid]:
                    end = mid - 1
                else:
                    start = mid + 1
            elif nums[mid] < nums[end]: # second half sorted
                if nums[mid] < target <= nums[end]:
                    start = mid + 1
                else:
                    end = mid - 1
            
            # start >= mid, mid >= end
            # 22222 0 1 2
            while start < len(nums) and nums[start] == nums[mid]:
                start += 1
            while end >= 0 and nums[end] == nums[mid]:
                end -= 1
                
        return False