数据结构-Reversing Linked List

题目

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10​5​​) which is the total number of nodes, and a positive K (≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

运行结果

Case	Hint	Result	Run Time	Memory
0	sample 有尾巴不反转, 地址取上下界	Accepted	2 ms	352 KB
1	正好全反转	Accepted	3 ms	352 KB
2	K=N全反转	Accepted	3 ms	416 KB
3	K=1不用反转	Accepted	2 ms	416 KB
4	N=1 最小case	Accepted	2 ms	384 KB
5	最大N,最后剩K-1不反转	Accepted	153 ms	6048 KB
6	有多余结点不在链表上	Accepted	3 ms	416 KB

总结：程序占用的内存稍多，还可以继续改进，有看到更好的运行结果：

http://www.pianshen.com/article/2800296730/

程序

#include
using namespace std;

typedef struct node *PtrToNode;
struct node
{
    int adress;
    int data;
    int nextAdress;
    PtrToNode next;
};
typedef PtrToNode List;

int MAX_ADRESS = 100000;
//真实链表长度
int TrueN = 0;

List read(int firstAdress, int N, int K);
void printList(List L);
List read_test(int firstAdress, int N, int K);
List reverseKsublist(List L, int N, int K);


int main()
{
    List L, newL;
    int firstAdress, N, K;

    cin >> firstAdress;
    cin >> N;
    cin >> K;
    L = read(firstAdress, N, K);

    /* 测试用例
    firstAdress = 100;
    N = 7;
    K = 7;
    L = read_test(firstAdress, N, K);
    */
    newL = reverseKsublist(L, TrueN, K);
    printList(newL);
    return 0;
}

List reverSubList(List subList, int K)
{
    List FirstToL, SecToL, ThreeToL;

    FirstToL = subList->next;
    SecToL = FirstToL->next;
    ThreeToL = NULL;

    for(int i=0; inext;
        SecToL->next = FirstToL;
        SecToL->nextAdress = FirstToL->adress;
        //三个指针后移
        FirstToL = SecToL;
        SecToL = ThreeToL;

    }
    //将反转后的最后一个元素接到下一个链表第一个元素
    subList->next->next = SecToL;
    if(ThreeToL){
        subList->next->nextAdress = SecToL->adress;
    }
    else{
        subList->next->nextAdress = -1;
    }

    subList->next = FirstToL;
    subList->nextAdress = FirstToL->adress;

    return subList;
}

List reverseKsublist(List L, int N, int K)
{
    List newSubList, temp;

    if(K == 1){return L;}
    if(K>N){K=N;}

    List subListHead = L;
    for(int i=0; inext;}

    }
    return L;

}

//测试用例
List read_test(int firstAdress, int N, int K)
{
    int data[MAX_ADRESS];
    int next[MAX_ADRESS];
    int tempAdr;
    List L, tempL, temp;

    L = new node;
    L->nextAdress = firstAdress;
    L->next = NULL;
    tempL = L;

    //以数据地址作为数组的索引
    //1
    tempAdr = 0;
    data[tempAdr] = 4;
    next[tempAdr] = 99999;
    //2
    tempAdr = 100;
    data[tempAdr] = 1;
    next[tempAdr] = 12309;
    //3
    tempAdr = 68237;
    data[tempAdr] = 6;
    next[tempAdr] = -1;
    //4
    tempAdr = 33218;
    data[tempAdr] = 3;
    next[tempAdr] = 0;
    //5
    tempAdr = 99999;
    data[tempAdr] = 5;
    next[tempAdr] = 68237;
    //6
    tempAdr = 12309;
    data[tempAdr] = 2;
    next[tempAdr] = 33218;
    //7 ->多余结点，不会出现在链表上的
    tempAdr = 00002;
    data[tempAdr] = 10;
    next[tempAdr] = 25468;

    //构建链表结构
    while(1)
    {
        ++TrueN;
        temp = new node;
        if(next[tempL->nextAdress] == -1){
            temp->adress = tempL->nextAdress;
            temp->data = data[temp->adress];
            temp->nextAdress = next[temp->adress];
            temp->next = NULL;
            tempL->next = temp;

            tempL = tempL->next;
            tempL->next = NULL;
            break;
        }
        temp->adress = tempL->nextAdress;
        temp->data = data[temp->adress];
        temp->nextAdress = next[temp->adress];
        tempL->next = temp;
        tempL = tempL->next;

    }

    //printList(L);

    return L;
}

List read(int firstAdress, int N, int K)
{
    int data[MAX_ADRESS];
    int next[MAX_ADRESS];
    int tempAdr;
    List L, tempL, temp;

    L = new node;
    L->nextAdress = firstAdress;
    L->next = NULL;
    tempL = L;

    //以数据地址作为数组的索引
    for(int i=0; i> tempAdr;
        cin >> data[tempAdr];
        cin >> next[tempAdr];
    }

    //构建链表结构
    while(1)
    {
        ++TrueN;
        temp = new node;
        if(next[tempL->nextAdress] == -1){
            temp->adress = tempL->nextAdress;
            temp->data = data[temp->adress];
            temp->nextAdress = next[temp->adress];
            temp->next = NULL;
            tempL->next = temp;

            tempL = tempL->next;
            tempL->next = NULL;
            break;
        }
        temp->adress = tempL->nextAdress;
        temp->data = data[temp->adress];
        temp->nextAdress = next[temp->adress];
        tempL->next = temp;
        tempL = tempL->next;
    }

    //printList(L);

    return L;
}

void printList(List L)
{
    List tempL;

    tempL = L->next;
    while(tempL)
    {
        if(tempL->nextAdress == -1)
            printf("%.5d %d %d\n", tempL->adress, tempL->data, tempL->nextAdress);
        else
            //格式输出，%.5意味着如果一个整数不足5位，输出时前面补0 如：22，输出：00022
            printf("%.5d %d %.5d\n", tempL->adress, tempL->data, tempL->nextAdress);
        tempL = tempL->next;
    }
}