def removeElements(self, head: ListNode, val: int) -> ListNode:

# If head is None or the val is not in head, return head
if not head:
return head
if head.val == val:
head = head.next
return self.removeElements(head, val)

# Traverse through the list
prev = None
curr = head
while curr:
if curr.val == val:
if prev:
prev.next = curr.next
else:
head = curr.next
curr = curr.next
else:
prev = curr
curr = curr.next

return head

相关问题

def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:

`removeElements` 函数用于从给定的链表（`head`）中删除所有值等于 `val` 的元素，返回新链表的头节点。这个函数采用迭代方式实现：

1. 创建一个虚拟头节点 `dummy`，并将其 `next` 指向原始链表的头 `head`。
2. 使用一个指针 `cur` 初始化为 `dummy`，然后开始遍历链表。
3. 在每次循环中，检查当前节点 `cur.next` 的值是否等于 `val`。如果相等，则跳过该节点，通过设置 `cur.next` 为 `cur.next.next` 来移除它。
4. 如果不等于 `val`，则简单地移动 `cur` 到下一个节点。
5. 循环结束后，`dummy.next` 就指向了新的链表头，所以返回 `dummy.next`。

代码示例如下[^1]:

def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
    dummy = ListNode()  # 创建虚拟头节点
    dummy.next = head  # 将 dummy 的 next 指向原头

    cur = dummy  # 定义当前节点
    while cur and cur.next:  # 当还有节点时继续
        if cur.next.val == val:  # 找到要删除的节点
            cur.next = cur.next.next  # 跳过它
        else:
            cur = cur.next  # 否则，移动到下一个节点

    return dummy.next  # 返回新链表的头节点

用c语言写class ListNode: def __init__(self, x): self.val = x self.next = Nonedef mergeTwoLists(l1: ListNode, l2: ListNode) -> ListNode: if not l1: return l2 if not l2: return l1 if l1.val <= l2.val: head = l1 l1 = l1.next else: head = l2 l2 = l2.next cur = head while l1 and l2: if l1.val <= l2.val: cur.next = l1 l1 = l1.next else: cur.next = l2 l2 = l2.next cur = cur.next if l1: cur.next = l1 else: cur.next = l2 return head# 测试l1 = ListNode(1)l1.next = ListNode(3)l1.next.next = ListNode(5)l2 = ListNode(2)l2.next = ListNode(4)l2.next.next = ListNode(6)head = mergeTwoLists(l1, l2)while head: print(head.val, end=' ') head = head.next

以下是使用C语言实现的代码：

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){
    if (!l1) return l2;
    if (!l2) return l1;

    struct ListNode *head, *cur;
    if (l1->val <= l2->val) {
        head = l1;
        l1 = l1->next;
    } else {
        head = l2;
        l2 = l2->next;
    }
    cur = head;

    while (l1 && l2) {
        if (l1->val <= l2->val) {
            cur->next = l1;
            l1 = l1->next;
        } else {
            cur->next = l2;
            l2 = l2->next;
        }
        cur = cur->next;
    }

    if (l1) {
        cur->next = l1;
    } else {
        cur->next = l2;
    }

    return head;
}

int main() {
    struct ListNode *l1, *l2, *head;
    l1 = (struct ListNode *)malloc(sizeof(struct ListNode));
    l1->val = 1;
    l1->next = (struct ListNode *)malloc(sizeof(struct ListNode));
    l1->next->val = 3;
    l1->next->next = (struct ListNode *)malloc(sizeof(struct ListNode));
    l1->next->next->val = 5;
    l1->next->next->next = NULL;

    l2 = (struct ListNode *)malloc(sizeof(struct ListNode));
    l2->val = 2;
    l2->next = (struct ListNode *)malloc(sizeof(struct ListNode));
    l2->next->val = 4;
    l2->next->next = (struct ListNode *)malloc(sizeof(struct ListNode));
    l2->next->next->val = 6;
    l2->next->next->next = NULL;

    head = mergeTwoLists(l1, l2);
    while (head) {
        printf("%d ", head->val);
        head = head->next;
    }
    printf("\n");

    return 0;
}

输出结果为：1 2 3 4 5 6