02-线性结构2 Reversing Linked List

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反转链表的练习：下面是自己试的代码

#include <stdio.h>#include <stdlib.h>typedef struct Node{    int Address;    int data;    int nextAddress;    struct Node *next;}Node;typedef struct Node *LinkList;int main(){    LinkList L1 , p1 , q1;    int i;    int firstAddress , N , K;    L1 = (LinkList)malloc(sizeof(Node));    L1->next = NULL;    scanf("%d %d %d",&firstAddress,&N,&K);    p1 = L1;    for(i=0;i<N;i++){        q1 = (LinkList)malloc(sizeof(Node));        scanf("%d %d %d",&q1->Address,&q1->data,&q1->nextAddress);        p1->next = q1;        p1 = q1;    }    p1->next = NULL;    printf("测试: \n");    p1 = L1->next;    while(p1){        printf("%d %d %d\n",p1->Address,p1->data,p1->nextAddress);        p1 = p1->next;    }    //sort the linklist    LinkList L2 , p2 , q2;    L2 = (LinkList)malloc(sizeof(Node));    L2->next = NULL;    p2 = L2;    int findAddress;    findAddress= firstAddress;    while(findAddress != -1){        q2 = (LinkList)malloc(sizeof(Node));        p1 = L1;        while(p1 && findAddress != p1->Address){            p1 = p1->next;        }        q2->Address = p1->Address;        q2->data = p1->data;        q2->nextAddress = p1->nextAddress;        findAddress = p1->nextAddress;        p2->next = q2;        p2 = q2;    }    p2->next = NULL;    printf("测试2: \n");    p2 = L2->next;    while(p2){        printf("%d %d %d\n",p2->Address,p2->data,p2->nextAddress);        p2 = p2->next;    }    //start to reserve list    int cnt = 1;    LinkList New , Old , temp;    New = L2->next;    Old = New->next;    while(cnt < K){        temp = Old->next;        Old->next = New;        New = Old;        Old = temp;        cnt++;    }    L2->next->next = Old;    printf("反转的链表：\n");    while(New){        printf("%d %d %d\n",New->Address,New->data,New->nextAddress);        New = New->next;    }}

反转链表的原理用陈越老师的图,说明一下：

不过我没想明白，如果K反转的个数是链表的单元个数呢，比如说6。。。

最后没通过。。。。。。。。

然后

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

typedef struct Node
{
    int address;
    int data;
    int nextAddress;
    struct Node *next;
}Node;
typedef struct Node *LinkList;

int main()
{
    //排序前
    LinkList L1, p1, q1;
    L1 = (LinkList)malloc(sizeof(Node));    //创建头指针
    L1->next = NULL;
    int firstAddress;
    int N, K;//N为总结点数 K为需翻转的数
    scanf("%d %d %d", &firstAddress, &N, &K);
    p1 = L1;
    for(int i = 0; i < N; i++) {
        q1 = (LinkList)malloc(sizeof(Node));
        scanf("%d %d %d",&q1->address, &q1->data, &q1->nextAddress);
        p1->next = q1;
        p1 = q1;
    }
    p1->next = NULL;

//    //测试没问题
//    printf("测试1 ：\n");
//    p1 = L1->next;
//    while(p1){
//        printf("%05d %d %d\n", p1->address, p1->data, p1->nextAddress);
//        p1 = p1->next;
//    }

    //排序后
    LinkList L2, p2;
    L2 = (LinkList)malloc(sizeof(Node));    //创建头指针
    L2->next = NULL;
    int count = 0;
    int findAddress = firstAddress;
    p2 = L2;
    while(findAddress != -1) {            //while(count < N) {有多余结点不在链表上没通过

        q1 = L1;
        while(q1->next) {
            if(q1->next->address == findAddress) {
                p2->next = q1->next;
                q1->next = q1->next->next;
                p2 = p2->next;
                count++;
//                printf("count = %d\n",count);
                findAddress = p2->nextAddress;
//                printf("findAddress = %d\n",findAddress);
            }else {
                q1 = q1->next;
            }
        }
    }
    p2->next = NULL;

//    //测试没问题
//    printf("测试2 ：\n");
//    p2 = L2->next;
//    while(p2){
//        printf("%05d %d %05d\n", p2->address, p2->data, p2->nextAddress);
//        p2 = p2->next;
//    }
    //Reversing
    LinkList L3, p3, q3, tail;
    L3 = (LinkList)malloc(sizeof(Node));    //创建头指针
    L3->next = NULL;
    //将L2以头插法插入L3
    int n = count;                //防止有多余结点影响 n=N 会影响
    int k = K;
    p3 = L3;
    p2 = L2;
    while(n >= k) {
        n -= k;
        for(int i = 0; i < k; i++) {
            p3->next = p2->next;
            p2->next = p2->next->next;
            if(i == 0)
                tail = p3->next;
            else
                p3->next->next = q3;
            q3 = p3->next;
        }
        p3 = tail;
    }
    p3->next = L2->next;

    p3 = L3->next;
    while(p3->next) {
        printf("%05d %d %05d\n",p3->address, p3->data, p3->next->address);//不到五位数用0补全
        p3 = p3->next;
    }
    printf("%05d %d -1\n",p3->address, p3->data);
    return 0;
}

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