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這題是33的follow-up 如果多了重複怎麼辦?

Problem

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 = 0Output: true

Example 2:

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

 

Constraints:

  • 1 <= nums.length <= 5000
  • -104 <= nums[i] <= 104
  • nums is guaranteed to be rotated at some pivot.
  • -104 <= target <= 104

 

Follow up: This problem is similar to Search in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

Sol

遇到重複就把重複的去掉就好，把與中點重複的去掉就好

def f(ns,n,l,r):
    #print(l,r)
    if abs(r-l) <= 1:
        return ns[l] == n or ns[r] == n
    else:
        mid = (l+r)//2
        if n == ns[mid]:
            return True
        elif ns[l] < ns[mid]:  
            if ns[l] <= n and n <= ns[mid]:
                return f(ns,n,l,mid)
            else:
                return f(ns,n,mid+1,r)
        elif ns[mid] < ns[r]:
            if n >= ns[mid] and n <= ns[r]:
                return f(ns,n,mid,r)
            else:
                return f(ns,n,l,mid-1)
        elif ns[l] == ns[mid]:
            return f(ns,n,l+1,r)
        else:
            return f(ns,n,l,r-1)
            
        
class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if not nums:
            return False
        return f(nums,target,0,len(nums)-1)