我如何排序中的一个字段结构的链接列表？ [英] How do I sort a linked list of structures by one of the fields?

问题描述

哇现在我知道我不。哈哈。

Wow now i know I dont. Lol.

我有我的结构是这样的：

I've got my structure like this:

struct Medico{ 
int Id_Doctor;
int Estado;
char Nombre[60]; ////focus on this part of the structure, this is name.
char Clave_Acceso[20];
char Especialidad[40];
struct Medico *next;
};

和我想组织结构取决于名称（按字母顺序排列..）关于如何解决这个问题的任何想法？

And I want to organize the structure depending on the name(alphabetical order..) any ideas on how to tackle this problem?

例如：

Albert Haynesworth
Bob Marley
Carl Johnson

感谢您非常先进。 ：）（C，UNIX）

Thank you very much in advanced. :)(C, Unix)

推荐答案

实施了在C链表一个合并是很容易的：

Implementing a mergesort over a linked list in C is quite easy:

#include <string.h>
#include <stdlib.h>
#include <stdio.h>

struct node {
    struct node *next;
    char *data;
};

struct node *
divlist (struct node *n) {
    int i = 0;
    if (n) {
        struct node *tail, *n2 = n;
        while (1) {
            n2 = n2->next;
            if (!n2) break;
            if (i++ & 1) n = n->next;
        }
        tail = n->next;
        n->next = NULL;
        return tail;
    }
    return NULL;
}

struct node *
mergelists(struct node *a, struct node *b) {
    struct node *n;
    struct node **last = &n;
    if (!a) return b;
    if (!b) return a;

    while (1) {
        if (strcmp(a->data, b->data) > 1) {
            *last = b;
            last = &b->next;
            b = b->next;
            if (!b) {
                *last = a;
                break;
            }
        }
        else {
            *last = a;
            last = &a->next;
            a = a->next;
            if (!a) {
                *last = b;
                break;
            }
        }
    }
    return n;
}

struct node *
sortlist (struct node *n) {
    struct node *tail = divlist(n);
    if (!tail) return n;
    return mergelists(sortlist(n), sortlist(tail));
}

int main(int argc, char *argv[]) {
    int i;
    struct node *n1, *n = NULL;
    for (i = argc; --i >= 1;) {
        n1 = (struct node *)malloc(sizeof(*n1));
        n1->data = argv[i];
        n1->next = n;
        n = n1;
    }

    n1 = n = sortlist(n);

    while (n1) {
        printf("%s\n", n1->data);
        n1 = n1->next;
    }
    return 0;
}

请注意，你将不得不修改此code使用你的数据结构和正确的比较！

Note that you will have to modify this code to use your data structure and the right comparison!

这篇关于我如何排序中的一个字段结构的链接列表？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！