【经典LeetCode算法题目专栏分类】【第7期】快慢指针与链表
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快慢指针
移动零
| class Solution: def moveZeroes(self, nums: List[int]) -> None: """ Do not return anything, modify nums in-place instead. """ left = 0 n = len(nums) for i in range(n): if nums[i] != 0: nums[left],nums[i] = nums[i],nums[left] left += 1 return nums class Solution: def moveZeroes(self, nums: List[int]) -> None: """ Do not return anything, modify nums in-place instead. """ j = 0 for i in range(len(nums)): if nums[i] != 0: nums[j] = nums[i] if i != j: nums[i] = 0 j += 1 return nums |
链表
两两交换链表中的节点
| # 迭代 class Solution: def swapPairs(self, head: ListNode) -> ListNode: # 通过迭代实现 dummy = ListNode(-1) dummy.next = head prev_node = dummy while head and head.next: first_node = head second_node = head.next # 交换节点 prev_node.next = second_node first_node.next = second_node.next second_node.next = first_node # 初始化头节点与prev_node prev_node = first_node head = first_node.next return dummy.next # 递归 class Solution: def swapPairs(self, head: ListNode) -> ListNode: # 递归实现 if not head or not head.next: return head first_node = head second_node = head.next # 第二个节点的next节点作为头部传入递归函数，返回的是 # 指向第二个节点的指针 first_node.next = self.swapPairs(second_node.next) second_node.next = first_node return second_node |
反转链表
将链表进行反转
| # 迭代 class Solution: def reverseList(self, head: ListNode) -> ListNode: if head is None: return head pre = None cur = head while cur: nxt = cur.next cur.next = pre pre = cur cur = nxt return pre # 递归 class Solution: def reverseList(self, head: ListNode) -> ListNode: if not head or not head.next: return head last = self.reverseList(head.next) head.next.next = head head.next = None return last |
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