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这是 拼题A(PTA)《中M2019秋C入门和进阶练习集》的习题。原题在https://pintia.cn/problem-sets/1163286449659043840/problems/1174288506294865941 (侵删)
本人的答案仅供交流学习,请勿用于当作答案来提交!
题目描述
6-22 学生成绩链表处理 (20 point(s))
本题要求实现两个函数,一个将输入的学生成绩组织成单向链表;另一个将成绩低于某分数线的学生结点从链表中删除。
函数接口定义:
struct stud_node *createlist();
struct stud_node *deletelist( struct stud_node *head, int min_score ); 函数createlist利用scanf从输入中获取学生的信息,将其组织成单向链表,并返回链表头指针。链表节点结构定义如下: struct stud_node {
int num; //学号
char name[20]; //姓名
int score; //成绩
struct stud_node *next; //指向下个结点的指针
};
输入为若干个学生的信息(学号、姓名、成绩),当输入学号为0时结束。
函数deletelist从以head为头指针的链表中删除成绩低于min_score的学生,并返回结果链表的头指针。
裁判测试程序样例:
#include
#include
struct stud_node {
int num;
char name[20];
int score;
struct stud_node *next;
};
struct stud_node *createlist();
struct stud_node *deletelist( struct stud_node *head, int min_score );
int main()
{
int min_score;
struct stud_node *p, *head = NULL;
head = createlist();
scanf("%d", &min_score);
head = deletelist(head, min_score);
for ( p = head; p != NULL; p = p->next )
printf("%d %s %d\n", p->num, p->name, p->score);
return 0;
}
// 你的代码将被嵌在这里
输入样例:
1 zhang 78
2 wang 80
3 li 75
4 zhao 85
0
80
输出样例:
2 wang 80
4 zhao 85
我的答案
/*================================================================
* Copyright (C) 2019 程序知路. All rights reserved.
*
* Filename :6-22-学生成绩链表处理.c
* Author :程序知路
* E-Mail :admin@icxzl.com
* Create Date :2019年10月24日
* Description :
================================================================*/
#include
#include
struct stud_node {
int num;
char name[20];
int score;
struct stud_node *next;
};
struct stud_node *createlist();
struct stud_node *deletelist( struct stud_node *head, int min_score );
int main()
{
int min_score;
struct stud_node *p, *head = NULL;
head = createlist();
// for ( p = head; p != NULL; p = p->next )
// printf("%d %s %d\n", p->num, p->name, p->score);
// return 0;
scanf("%d", &min_score);
// printf("%d\n", min_score);
// return 0;
head = deletelist(head, min_score);
for ( p = head; p != NULL; p = p->next )
printf("%d %s %d\n", p->num, p->name, p->score);
return 0;
}
// 以下是有效代码
struct stud_node* createlist() {
int num;
char name[20];
int score;
struct stud_node *head = NULL, *current = NULL, *prev = NULL;
while (scanf("%d %s %d", &num, name, &score) == 3) {
size_t i;
if (!head) {
head = (struct stud_node*) malloc(sizeof(struct stud_node));
head->num = num;
for ( i = 0; i name[i] = name[i];
}
head->name[i] = '\0';
head->score = score;
current = head;
} else {
current = (struct stud_node*) malloc(sizeof(struct stud_node));
current->num = num;
for (i = 0; i name[i] = name[i];
}
current->name[i] = '\0';
current->score = score;
prev->next = current;
}
prev = current;
}
if (head)
current->next = NULL;
size_t len = 0;
size_t i;
for (i = 0; i 0; -- i)
ungetc(name[i], stdin);
ungetc(name[i], stdin);
return head;
}
struct stud_node *deletelist( struct stud_node *head, int min_score ) {
// 解决方案1
//struct stud_node *p = NULL, *current = NULL, *prev = NULL;
// for ( p = head; p != NULL; p = p->next )
// printf("%d %s %d\n", p->num, p->name, p->score);
// return head;
// for (; head != NULL; head = head->next) {
// // printf("%s\n", head->name);
// // continue;
// if (head->score >= min_score) {
// if (p == NULL) {
// // printf("\n====\n");
// p = head;
// current = p;
// } else {
// // printf("\n+++\n");
// current = head;
// prev->next = current;
// }
// prev = current;
// }
// }
//
// if (p)
// current->next = NULL;
//
// return p;
// 解决方案2
struct stud_node *p = head, *prev = NULL;
for (; p != NULL; p = p->next) {
if (p->score next;
} else {
prev->next = p->next;
}
} else {
prev = p;
}
}
return head;
}
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