本題要求實現一個合並兩個有序鏈表的簡單函數。鏈表結點定義如下:
struct ListNode { int data; struct ListNode *next; };
函數接口定義:
struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2);
其中list1
和list2
是用戶傳入的兩個按data
升序鏈接的鏈表的頭指針;函數mergelists
將兩個鏈表合並成一個按data
升序鏈接的鏈表,並返回結果鏈表的頭指針。
裁判測試程序樣例:
#include <stdio.h> #include <stdlib.h> struct ListNode { int data; struct ListNode *next; }; struct ListNode *createlist(); /*裁判實現,細節不表*/ 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; } /* 你的代碼將被嵌在這里 */
輸入樣例:
1 3 5 7 -1 2 4 6 -1
輸出樣例:
1 2 3 4 5 6 7
struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2) { int len = 0; int a[10000]; struct ListNode *p1 = list1; struct ListNode *p2 = list2; struct ListNode *head = NULL; struct ListNode *tail = NULL; struct ListNode *q; while(p1) { a[len] = p1->data; p1 = p1->next; len++; } while(p2) { a[len] = p2->data; p2 = p2->next; len++; } int i, j, temp; for(i=1; i<len; i++) { for(j=0; j<len-i; j++) { if(a[j]>a[j+1]) { temp = a[j+1]; a[j+1] = a[j]; a[j] = temp; } } } for(i = 0; i < len; i++) { q = (struct ListNode *)malloc(sizeof(struct ListNode)); q->data = a[i]; if(head == NULL) { head = q; head->next = NULL; } if(tail != NULL)//tail為開辟節點 { tail->next = q; } tail = q; tail->next = NULL; } return head; }