l1=list(map(int,input().split(' '))) l2=list(map(int,input().split(' '))) a=list(set(l1+l2)) a.sort() str1=' '.join(a) print(str1)查看错误并修改
时间: 2024-05-21 10:10:37 浏览: 13
l1 = list(map(int, input().split(' ')))
l2 = list(map(int, input().split(' ')))
a = list(set(l1 + l2))
a.sort()
str1 = ' '.join(map(str, a))
print(str1)
相关问题
# 请在这里填写答案 L1 = map(int, input().split(" ")) L2 = map(int, input().split(" ")) try: print(get_ratios(L1, L2)) except Exception as e: print(e)
这段代码是一个 Python 程序,首先读取两行输入,每行输入都是一串数字,使用 `map` 函数将字符串类型的数字转换为整型,并存储在 `L1` 和 `L2` 两个列表中。然后调用 `get_ratios` 函数,该函数接受两个参数,均为列表类型,用于计算两个列表对应位置上的数的比例,并将结果存储在一个新的列表中。最后使用 `try...except` 语句来捕获可能出现的异常,如果出现异常则打印异常信息。但是这段代码中缺少 `get_ratios` 函数的实现,所以无法运行。
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