1 问题
合并2个已经排好序的链接，比如
1->3->5->7
2->4->6
合并后新的链表如下
1->2->3->4->5->6->7
2 代码实现
#include <stdio.h>
typedef struct Node
{
int val;
struct Node *next;
} Node;
/*
*print list
*/
void print_list(Node *head)
{
if (head == NULL)
{
printf("head is NULL\n");
return;
}
Node *p = head;
while (p != NULL)
{
printf("value is %d\n", p->val);
p = p->next;
}
}
/*
*合并链表
*/
struct Node* merge(Node *head1, Node *head2)
{
if (head1 == NULL)
{
return head2;
}
if (head2 == NULL)
{
return head1;
}
struct Node *new = NULL;
if (head1->val < head2->val)
{
new = head1;
new->next = merge(head1->next, head2);
}
else
{
new = head2;
new->next = merge(head1, head2->next);
}
return new;
}
int main()
{
//list1 0->3->5->9;
Node head, node1, node2, node3;
head.val = 0;
head.next = &node1;
node1.val = 3;
node1.next = &node2;
node2.val = 5;
node2.next = &node3;
node3.val = 9;
node3.next = NULL;
printf("list1 is such as\n");
print_list(&head);
//list2 1->4->6
Node head1, node4, node5;
head1.val = 1;
head1.next = &node4;
node4.val = 4;
node4.next = &node5;
node5.val = 6;
node5.next = NULL;
printf("list2 is such as\n");
print_list(&head1);
printf("merge list1 and list2\n");
Node *new = merge(&head, &head1);
print_list(new);
return 0;
}
3 运行结果
list1 is such as value is 0 value is 3 value is 5 value is 9 list2 is such as value is 1 value is 4 value is 6 merge list1 and list2 value is 0 value is 1 value is 3 value is 4 value is 5 value is 6 value is 9
4 总结
我一开始写成这样了
struct Node* merge(Node *head1, Node *head2)
{
if (head1 == NULL)
{
return head2;
}
if (head2 == NULL)
{
return head1;
}
struct Node *new = NULL;
while (head1 != NULL && head2 != NULL)
{
if (head1->val < head2->val)
{
new = head1;
new->next = merge(head1->next, head2);
}
else
{
new = head2;
new->next = merge(head1, head2->next);
}
}
return new;
}
加了while循环 又是递归，肯定容易出问题，一定要记住，递归函数里面循环体里面又有递归一般就有问题
