1025 反转链表 (25 分)

自己AC代码：

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

typedef struct {
    int ad;
    int data;
    int next;
    int sq;
}Node;

int cmp1(const void *a,const void *b){
    Node *pa=(Node*)a;
    Node *pb=(Node*)b;
    if(pa->sq<pb->sq)
        return -1;
    else
        return 1;
}
int cmp2(const void *a,const void *b){
    Node *pa=(Node*)a;
    Node *pb=(Node*)b;
    if(pa->sq<pb->sq)
        return 1;
    else
        return -1;
}
void change(Node *a,int n){
    int first=a[0].next,i;
    for(i=0;i<n-1;i++){
        a[i].next=a[i+1].ad;
    }
    a[i].next=first;
}
int main(){
    int fa,n,k,i=0,sq=1,sum=0;
    scanf("%d %d %d",&fa,&n,&k);
    Node node[100010]={0};
    int find;
    int x=n;
    while(x--){
        scanf("%d %d %d",&node[i].ad,&node[i].data,&node[i].next);
        if(node[i].ad==fa){
            find=node[i].next;
            node[i].sq=sq++;
            sum++;
        }
        i++;
    }
    int j=0;
    while(find!=-1){
        if(node[j].ad==find){
            node[j].sq=sq++;
            find=node[j].next;
            sum++;
        }
        j++;
        if(j==n)
            j=0;
    }

    qsort(node,n,sizeof(Node),cmp1);

    for(int v=0;v<sum/k;v++){
        qsort(node+n-sum+v*k,k,sizeof(Node),cmp2);
    }
    for(int v=0;v<sum/k;v++){
        change(node+n-sum+v*k,k);
    }
    for(int v=n-sum+k-1;v<n;v+=k){
        node[v].next=node[v+1].ad;
    }
    for(int v=n-sum;v<n;v++){
        if(v<n-1)
            printf("%05d %d %05d\n",node[v].ad,node[v].data,node[v].next);
        else
            printf("%05d %d -1\n",node[v].ad,node[v].data);
    }
    return 0;
}

柳神代码：

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
    int first, k, n, temp;
    cin >> first >> n >> k;
    int data[100005], next[100005], list[100005];
    for (int i = 0; i < n; i++) {
        cin >> temp;
        cin >> data[temp] >> next[temp];
    }
    int sum = 0;//不一定所有的输入的结点都是有用的，加个计数器
    while (first != -1) {
        list[sum++] = first;
        first = next[first];
    }
    for (int i = 0; i < (sum - sum % k); i += k)
        reverse(begin(list) + i, begin(list) + i + k);
    for (int i = 0; i < sum - 1; i++)
        printf("%05d %d %05d\n", list[i], data[list[i]], list[i + 1]);
    printf("%05d %d -1", list[sum - 1], data[list[sum - 1]]);
    return 0;
}

标签：1025,node,25,int,sum,list,Node,链表,next
来源： https://blog.csdn.net/Jay98/article/details/113836477