浙大数据结构课后习题练习二 7-2 Reversing Linked List (25 分)

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 (≤) which is the total number of nodes, and a positive K (≤) 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

#include <iostream>

#include <vector>

#define M 100000

using namespace std;

struct ele{

    int addr;

    int data;

    int next=-;

};

int main(){

    int iniAddr,num,revNum,tmpAddr,tmpData,tmpNext;

    ele element[M],last;

    cin>>iniAddr>>num>>revNum;

    while(num--){

        cin>>tmpAddr>>tmpData>>tmpNext;

        element[tmpAddr].addr=tmpAddr;

        element[tmpAddr].data=tmpData;

        element[tmpAddr].next=tmpNext;

    }

    vector<ele> vec;

    while(iniAddr!=-){

        vec.push_back(element[iniAddr]);

        iniAddr=element[iniAddr].next;

    }

    //反转

    int point=;bool start=true;

    while(true){

        point+=revNum;

        if(point>vec.size()) break;

        else{

            for(int i=point-,j=;j<revNum;j++,i--){

                if(start) printf("%05d ",vec[i].addr);

                else printf("%05d\n%05d ",vec[i].addr,vec[i].addr);

                printf("%d ",vec[i].data);

                last=vec[i];

                start=false;

            }

        }

    }

    point-=revNum;

    for(int i=point;i<vec.size();i++){

        if(start) printf("%05d ",vec[i].addr);

        else printf("%05d\n%05d ",vec[i].addr,vec[i].addr);

        printf("%d ",vec[i].data);

        start=false;

        last=vec[i];

    }

    printf("-1");

    system("pause");

    return ;

}

巴特西

浙大数据结构课后习题练习二 7-2 Reversing Linked List (25 分)

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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巴特西

浙大数据结构课后习题 练习二 7-2 Reversing Linked List (25 分)

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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浙大数据结构课后习题练习二 7-2 Reversing Linked List (25 分)