publicListNodesortList(ListNode head){ // 获链表的总长度 intlength =0; // 从链表的头节点开始访问 ListNode node = head; // 利用 while 循环,可以统计出链表的节点个数,即长度 while(node !=null) { length++; node = node.next; } // 在原链表的头部设
") result = result.Next }}在这里插入图片描述Python完整代码如下：# -*-coding:utf-8-*-classListNode:def__init__(self, val=, next=None):self.val = valself.next = nextdefmodified_list(nums, head): has = set(nums) # 使用集合来存储nums中的值 dummy = ListNode(next=head)...
class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next def reverseLinkedList(head): if head is None: return None if head.next is None: return head prev = None curr = head next = None while curr is not None: next = curr.next curr.next = prev...
func main() { head := &ListNode{Val: 1} head.Next = &ListNode{Val: 2} head.Next.Next = &ListNode{Val: 3} head.Next.Next.Next = &ListNode{Val: 4} printlnLinkNodeList(head) k := 6 fmt.Println("k =", k) fmt.Println("---") ret := rotateRight(head, k) printlnLinkNodeList...
3、Python 代码classSolution: defswapPairs(self, head: ListNode)-> ListNode: # 寻找递归终止条件 # 1、head 指向的结点为 null # 2、head 指向的结点的下一个结点为 null # 在这两种情况下,一个节点或者空节点无论怎么交换操作,都是原来的 head ...
```python class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next def reverse_linked_list(head): prev = None curr = head while curr: next_node = curr.next curr.next = prev prev = curr curr = next_node return prev # 测试 node1 = ListNode(1)...
在所有 Python3 提交中击败了52.19%的用户# Definition for singly-linked list.# class ListNode:# ...
dummy_head = ListNode(next=head) cur = dummy_head while cur.next != None: if cur.next.val == val: cur.next = cur.next.next else: cur = cur.next return dummy_head.next 不使用dummy_head class Solution: def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListN...
nums := []int{1, 2, 3} head := &ListNode{Val: 1, Next: &ListNode{Val: 2, Next: &ListNode{Val: 3, Next: &ListNode{Val: 4, Next: &ListNode{Val: 5}}} result := modifiedList(nums, head) for result != nil { fmt.Print(result.Val, " ") result = result.Next } } 1. 2...
node=ListNode(data) n=get(head,index-1) node.next=n.next n.next=node #遍历链表 def show(head): p=head while p: print p.val; p=p.next; #删除节点 def delete(head,index): n=get(head,index-1) n.next=n.next.next head=ListNode(1); #测试代码 ...