链表的有序插入

从小到大排序

根据指针获取当前id，并设置前指针，方便操作：

// test1107.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include "stdio.h"
#include "memory.h"
#include <string.h>
#include <stdlib.h>
#include "string.h"

typedef struct LINK{
    int id;
    LINK *next;
}Link;

typedef struct LL{
    Link *root;
    int num;
}Ll;

void insert(Ll *l,int num);

int main(){
    Ll *ll = (Ll *)malloc(sizeof(Ll));
    ll->num = 0;
    ll->root = NULL;
    insert(ll,5);
    insert(ll,2);
    insert(ll,20);
    insert(ll,1);
    insert(ll,100);
    Link *link = (Link *)malloc(sizeof(Link));
    printf("%d",ll->root->id);
    link = ll->root->next;
    while(link != NULL){
        printf("->%d",link->id);
        link = link->next;
    }
    getchar();
    return 0;
}

void insert(Ll *l,int num){
    Link * linkp = (Link *)malloc(sizeof(Link));
    Link * current =  (Link *)malloc(sizeof(Link));
    Link * new_id = (Link *)malloc(sizeof(Link));
    current = l->root;
    linkp = l->root;
    while(current != NULL && current->id<num){
        linkp = current;
        current = current->next;
    }
    new_id->id = num;
    new_id->next = current;
    if(linkp == NULL){
        l->root = new_id;
    }else if(num < l->root->id){
        l->root = new_id;
    }else{
        linkp->next = new_id;
    }
}

从大到小排序

// test1107.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include "stdio.h"
#include "memory.h"
#include <string.h>
#include <stdlib.h>
#include "string.h"

typedef struct LINK{
    int id;
    LINK *next;
}Link;

typedef struct LL{
    Link *root;
    int num;
}Ll;

void insert(Ll *l,int num);

int main(){
    Ll *ll = (Ll *)malloc(sizeof(Ll));
    ll->num = 0;
    ll->root = NULL;
    insert(ll,5);
    insert(ll,2);
    insert(ll,20);
    insert(ll,1);
    insert(ll,5);
    insert(ll,100);
    Link *link = (Link *)malloc(sizeof(Link));
    printf("%d",ll->root->id);
    link = ll->root->next;
    while(link != NULL){
        printf("->%d",link->id);
        link = link->next;
    }
    getchar();
    return 0;
}

void insert(Ll *l,int num){
    Link * linkp = (Link *)malloc(sizeof(Link));
    Link * current =  (Link *)malloc(sizeof(Link));
    Link * new_id = (Link *)malloc(sizeof(Link));
    current = l->root;
    linkp = l->root;
    while(current != NULL && current->id>num){
        linkp = current;
        current = current->next;
    }
    new_id->id = num;
    new_id->next = current;
    if(linkp == NULL){
        l->root = new_id;
    }else if(num > l->root->id){
        l->root = new_id;
    }else{
        linkp->next = new_id;
    }
}