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->6Node 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