解释代码:def mergeTwoLists(l1: ListNode, l2: ListNode) -> ListNode:
时间: 2023-05-16 14:07:16 浏览: 197
这是一个 Python 函数,函数名为 mergeTwoLists,接受两个参数 l1 和 l2,它们的类型都是 ListNode,返回值是一个 ListNode 类型的对象。这个函数的作用是将两个有序链表合并成一个新的有序链表。具体实现可以参考 LeetCode 上的题目“合并两个有序链表”。
相关问题
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: if not l1: return l2 if not l2: return l1 if l1.val < l2.val: l1.next = self.mergeTwoLists(l1.next, l2) return l1 else l2.next = self.mergeTwoLists(l1, l2.next) return l2
这是一个 Python 代码段,实现了将两个链表合并成一个升序链表的功能。它使用了递归算法来实现链表的合并,当其中一个链表为空时,直接将另外一个链表返回,否则比较两个链表头结点的值,将较小的那个作为合并后的链表的头结点,然后将剩余的链表递归合并到这个头结点后面。
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:
```c
#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
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