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

浙大数据结构课后习题 练习二 7-2 Reversing Linked List (25 分)
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大家好,我是架构君,一个会写代码吟诗的架构师。今天说一说浙大数据结构课后习题 练习二 7-2 Reversing Linked List (25 分),希望能够帮助大家进步!!!

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=-1;
};
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!=-1){
        vec.push_back(element[iniAddr]);
        iniAddr=element[iniAddr].next;
    }
    //反转
    int point=0;bool start=true;
    while(true){
        point+=revNum;
        if(point>vec.size()) break;
        else{
            for(int i=point-1,j=0;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 0;
}

 

转载于:https://www.cnblogs.com/littlepage/p/11375093.html

本文来源weixin_30595035,由架构君转载发布,观点不代表Java架构师必看的立场,转载请标明来源出处:https://javajgs.com/archives/29444

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