习题11-7 奇数值结点链表(20 分)
本题要求实现两个函数,分别将读入的数据存储为单链表、将链表中奇数值的结点重新组成一个新的链表。链表结点定义如下:
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| struct ListNode { int data; ListNode *next; };
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函数接口定义:
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| struct ListNode *readlist(); struct ListNode *getodd( struct ListNode **L );
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函数readlist
从标准输入读入一系列正整数,按照读入顺序建立单链表。当读到−1时表示输入结束,函数应返回指向单链表头结点的指针。
函数getodd
将单链表L
中奇数值的结点分离出来,重新组成一个新的链表。返回指向新链表头结点的指针,同时将L
中存储的地址改为删除了奇数值结点后的链表的头结点地址(所以要传入L
的指针)。
裁判测试程序样例:
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| #include <stdio.h> #include <stdlib.h>
struct ListNode { int data; struct ListNode *next; };
struct ListNode *readlist(); struct ListNode *getodd( struct ListNode **L ); void printlist( struct ListNode *L ) { struct ListNode *p = L; while (p) { printf("%d ", p->data); p = p->next; } printf("\n"); }
int main() { struct ListNode *L, *Odd; L = readlist(); Odd = getodd(&L); printlist(Odd); printlist(L);
return 0; }
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输入样例:
输出样例:
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| #include <stdio.h> #include <stdlib.h>
struct ListNode { int data; struct ListNode *next; };
struct ListNode *readlist(); struct ListNode *getodd( struct ListNode **L ); void printlist( struct ListNode *L ) { struct ListNode *p = L; while (p) { printf("%d ", p->data); p = p->next; } printf("\n"); }
int main() { struct ListNode *L, *Odd; L = readlist(); Odd = getodd(&L); printlist(Odd); printlist(L);
return 0; }
void PushBack(int X, struct ListNode **rear) { struct ListNode *TmpCell;
TmpCell = malloc(sizeof(struct ListNode));
TmpCell->data = X; TmpCell->next = NULL; (*rear)->next = TmpCell; *rear = TmpCell; }
struct ListNode *readlist() { struct ListNode *TmpCell, *head, *tail; int data;
head = tail = malloc(sizeof(struct ListNode)); head->next = NULL;
scanf("%d", &data); while (data != -1) { PushBack(data, &tail); scanf("%d", &data); }
TmpCell = head; head = head->next; free(TmpCell);
return head; }
struct ListNode *getodd(struct ListNode **L) { struct ListNode *head, *p, *oddhead, *oddtail, *TmpCell;
p = head = malloc(sizeof(struct ListNode)); head->next = NULL;
oddhead = oddtail = malloc(sizeof(struct ListNode)); oddhead->next = NULL;
p->next = *L; while (p->next != NULL) { if (p->next->data % 2 != 0) { PushBack(p->next->data, &oddtail); TmpCell = p->next; p->next = TmpCell->next; free(TmpCell); } else p = p->next; }
TmpCell = head; head = head->next; free(TmpCell); *L = head;
TmpCell = oddhead; oddhead = oddhead->next; free(TmpCell);
return oddhead; }
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