1.?81——搜索旋轉排序數組2
已知存在一個按非降序排列的整數數組?nums?，數組中的值不必互不相同。
在傳遞給函數之前，nums?在預先未知的某個下標?k（0 <= k < nums.length）上進行了旋轉，使數組變為?[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]（下標從 0 開始計數）。例如，[0,1,2,4,4,4,5,6,6,7]?在下標?5?處經旋轉后可能變為?[4,5,6,6,7,0,1,2,4,4]?。
給你旋轉后的數組?nums?和一個整數?target?，請你編寫一個函數來判斷給定的目標值是否存在于數組中。如果?nums?中存在這個目標值?target?，則返回?true?，否則返回?false?。
你必須盡可能減少整個操作步驟。
標簽：數組，二分查找
代碼：
class Solution:def search(self, nums: List[int], target: int) -> bool:left = 0right = len(nums) - 1while left <= right:mid =left + (right - left) // 2if nums[mid] == target:return Trueif nums[left] == nums[mid]:left = left + 1continueelif nums[left] < nums[mid]:if nums[left] <= target < nums[mid]:right = mid - 1else:left = mid + 1elif nums[left] > nums[mid]:if nums[mid] < target <= nums[right]:left = mid + 1else:right = mid - 1return False2.?82——刪除排序鏈表中的重復元素2
給定一個已排序的鏈表的頭?head?，?刪除原始鏈表中所有重復數字的節點，只留下不同的數字?。返回?已排序的鏈表?。
標簽：鏈表，雙指針
代碼：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:pointer1 = dummy = ListNode()pointer2 = headwhile pointer2:val1 = pointer2.valval2 = pointer2.next.val if pointer2.next else 0if val1 != val2:pointer1.next = pointer2pointer1 = pointer1.nextpointer2 = pointer2.nextelse:while pointer2 and pointer2.val == val1:pointer2 = pointer2.nextpointer1.next = pointer2return dummy.next3.?83——刪除排序鏈表中的重復元素
給定一個已排序的鏈表的頭?head?，?刪除所有重復的元素，使每個元素只出現一次?。返回?已排序的鏈表?。
標簽：鏈表
代碼：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:if not head:return headelse:dummy = pointer = ListNode()pointer.next = headpointer = pointer.nextwhile head:if pointer.val != head.val:pointer.next = headpointer = pointer.nexthead = head.nextelse:head = head.nextpointer.next = Nonereturn dummy.next
)
)
