已知存在一个按非降序排列的整数数组 nums
,数组中的值不必互不相同。
在传递给函数之前,nums
在预先未知的某个下标 k
(0 <= k < nums.length
)上进行了旋转,使数组变为 [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]
(下标从 0 开始计数)。例如,[0,1,2,4,4,4,5,6,6,7]
在下标 5
处经旋转后可能变为 [4,5,6,6,7,0,1,2,4,4]
。
给你旋转后的数组 nums
和一个整数 target
,请你编写一个函数来判断给定的目标值是否存在于数组中。如果 nums
中存在这个目标值 target
,则返回 true
,否则返回 false
。
你必须尽可能减少整个操作步骤。
标签:数组,二分查找
代码:
class Solution:
def search(self, nums: List[int], target: int) -> bool:
left = 0
right = len(nums) - 1
while left <= right:
mid =left + (right - left) // 2
if nums[mid] == target:
return True
if nums[left] == nums[mid]:
left = left + 1
continue
elif nums[left] < nums[mid]:
if nums[left] <= target < nums[mid]:
right = mid - 1
else:
left = mid + 1
elif nums[left] > nums[mid]:
if nums[mid] < target <= nums[right]:
left = mid + 1
else:
right = mid - 1
return False
给定一个已排序的链表的头 head
, 删除原始链表中所有重复数字的节点,只留下不同的数字 。返回 已排序的链表 。
标签:链表,双指针
代码:
# 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 = head
while pointer2:
val1 = pointer2.val
val2 = pointer2.next.val if pointer2.next else 0
if val1 != val2:
pointer1.next = pointer2
pointer1 = pointer1.next
pointer2 = pointer2.next
else:
while pointer2 and pointer2.val == val1:
pointer2 = pointer2.next
pointer1.next = pointer2
return dummy.next
给定一个已排序的链表的头 head
, 删除所有重复的元素,使每个元素只出现一次 。返回 已排序的链表 。
标签:链表
代码:
# 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 head
else:
dummy = pointer = ListNode()
pointer.next = head
pointer = pointer.next
while head:
if pointer.val != head.val:
pointer.next = head
pointer = pointer.next
head = head.next
else:
head = head.next
pointer.next = None
return dummy.next