cur=cur.next if cur.val==val2: tmp=ListNode(val1) tmp.next=cur.next cur.next=tmp

哦，您想问的是关于链表的操作吧？我可以理解为您是想把链表中值为val2的节点替换为值为val1的节点，并保持节点的相对位置不变。对吧？

如果是这样的话，您可以使用如下的代码实现：

while cur:
    if cur.val == val2:
        tmp = ListNode(val1)
        tmp.next = cur.next
        cur.next = tmp
        break
    cur = cur.next

其中，cur代表当前节点，val1代表要插入的新值，val2代表需要替换的旧值。代码中，我们遍历链表，找到值为val2的节点后，创建一个新节点tmp，并将cur.next赋值给它的next属性，将tmp赋值给cur.next属性，从而在链表中将值为val2的节点替换为了值为val1的节点。

希望我的回答能够帮助到您！

相关问题

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

class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next class Solution: def sortList(self, head: ListNode) -> ListNode: if not head or not head.next: return head slow, fast = head, head.next while fast and fast.next: slow = slow.next fast = fast.next.next mid, slow.next = slow.next, None left, right = self.sortList(head), self.sortList(mid) dummy = ListNode(0) cur = dummy while left and right: if left.val < right.val: cur.next, left = left, left.next else: cur.next, right = right, right.next cur = cur.next cur.next = left if left else right return dummy.next解释每一行代码

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head

这里定义了一个链表节点类 ListNode，以及一个排序链表的类 Solution。sortList 函数接受一个链表头节点，返回排序后的链表头节点。

如果链表为空或只有一个节点，则直接返回该链表，因为只有一个节点时已经是有序的。

        slow, fast = head, head.next
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        mid, slow.next = slow.next, None

这里使用快慢指针找到链表的中间节点，并将链表一分为二。slow 指针每次走一步，fast 指针每次走两步，当 fast 走到链表尾部时，slow 刚好走到链表中间。将链表分成左右两个部分，中间节点 mid 作为右半部分的起始节点，将 slow 的 next 指针设为 None，断开左右两部分的链接。

        left, right = self.sortList(head), self.sortList(mid)

递归调用sortList函数，对左半部分和右半部分分别进行排序。

        dummy = ListNode(0)
        cur = dummy
        while left and right:
            if left.val < right.val:
                cur.next, left = left, left.next
            else:
                cur.next, right = right, right.next
            cur = cur.next
        cur.next = left if left else right
        return dummy.next

将左右两个有序链表合并成一个有序链表。使用一个新链表 dummy 和指针 cur，比较 left 和 right 的节点值，将较小的节点接到 cur 后面，然后将 cur 指针后移一个节点。最后，将 cur 的 next 指针指向剩余的节点。

最后返回排序后的链表头节点。