1 問題

合并2個已經排好序的鏈接，比如

1->3->5->7

2->4->6

合并后新的鏈表如下

1->2->3->4->5->6->7

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2 代碼實現

#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;} }/**合并鏈表*/ 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; }

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3 運行結果

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

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4 總結

我一開始寫成這樣了

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循環?又是遞歸，肯定容易出問題，一定要記住，遞歸函數里面循環體里面又有遞歸一般就有問題

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總結

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