本题要求实现一个合并两个有序链表的简单函数。链表结点定义如下:
struct ListNode { int data; struct ListNode *next; };
struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2);
其中list1
和list2
是用户传入的两个按data
升序链接的链表的头指针;函数mergelists
将两个链表合并成一个按data
升序链接的链表,并返回结果链表的头指针。
#include <stdio.h> #include <stdlib.h> struct ListNode { int data; struct ListNode *next; }; struct ListNode *createlist(); /*裁判实现,细节不表*/ struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2); void printlist( struct ListNode *head ) { struct ListNode *p = head; while (p) { printf("%d ", p->data); p = p->next; } printf("\n"); } int main() { struct ListNode *list1, *list2; list1 = createlist(); list2 = createlist(); list1 = mergelists(list1, list2); printlist(list1); return 0; } /* 你的代码将被嵌在这里 */
1 3 5 7 -1 2 4 6 -1
1 2 3 4 5 6 7
struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2) { struct ListNode *p1 = list1, *p2 = list2, *r = NULL, *head = list1; while (p2 != NULL) { if (p1 != NULL) //将list2插入list1 { r = p1; if (p1->data > p2->data) //比最小的小 { head = p2; p2 = p2->next; //list2下一个 head->next = p1; p1 = head; } else { while (p1 != NULL) { if (p1->data > p2->data) { r->next = p2; //前趋指向该数 r = r->next; //r指向后趋 p2 = p2->next; //原节点遍历到下一个 r->next = p1; //r的后趋指向下一个数 break; } r = p1; p1 = p1->next; //遍历 } } } if (p1 == NULL) //直接插入至最后 { r->next = p2; r = p2; p2 = p2->next; } } return head; }