所以我试图对整数的链表进行排序,但似乎找不到正确的方式来做到这一点,我的想法是采取无序列表,找到它的最大值,并将其放入另一个列表中。由于我不相信我可以从原始列表中删除节点,而不是双向链接,因此我曾计划将列表1中的节点设置为零值,从而将其状态作为最大值删除。因为这个原因,我打算运行一个固定的次数,每次找到下一个最大的值,直到列表1全部为0,列表2是列表1曾经是的有序版本。我已经创建了一个函数来做到这一点,但它似乎没有工作,虽然我找不到问题。使用另一个列表对一个链表进行排序
功能
#include <stdio.h>
#include <stdlib.h>
#include "functions.h"
struct num_node *create(struct num_node *list, int x){
struct num_node *current;
if (list == NULL){
list = (struct num_node*)malloc(sizeof(struct num_node));
list->num = x;
list->next = NULL;
return(list);
}
else{
current = (struct num_node *)malloc(sizeof(struct num_node));
current->num = x;
current->next = list;
return(current);
}
}
void print_nums(struct num_node *list) {
struct num_node *current;
for (current = list; current != NULL; current = current->next)
printf("%d\t", current->num);
}
struct num_node *sort_nums(struct num_node *list1, struct num_node *list2){
struct num_node *current;
struct num_node *large = list1;
for (int i = 0; i < 25; i++){
for (current = list1; current != NULL; current = current->next){
if (current->num > large->num){
large = current;
}
}
create(list2, large->num);
large->num = 0;
return(list2);
}
}
int sum(struct num_node *list){
int total = 0;
struct num_node *current;
for (current = list; current != NULL; current = current->next){
total = total + current->num;
}
return total;
}
float average(struct num_node *list){
float total = 0;
float count = 0;
struct num_node *current;
for (current = list; current != NULL; current = current->next){
total = total + current->num;
count++;
}
return total/count;
}
主要
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include "functions.h"
int main(){
struct num_node *head = NULL;
struct num_node *new_head = NULL;
srand(time(NULL));
for (int i = 1; i <= 25; i++){
int x = rand() % 100;
head = create(head, x);
}
print_nums(head);
sort_nums(head, new_head);
printf("\n");
printf("\n");
print_nums(new_head);
printf("\n");
printf("\n");
printf("The total of all numbers is: ");
printf("\t%d\n", sum(new_head));
printf("The average of the numbers is: ");
printf("\t%.3f\n", average(new_head));
}
“因为我不相信我可以删除从原来的列表中的节点而不会被双重链接”其实你可以删除单链表中的节点过多,你就必须有一个参考其前任。 – 2015-02-05 22:23:04