1074. Reversing Linked List (25)-白红宇

1074. Reversing Linked List (25)

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发布时间：2019-06-09

本文共 2864 字，大约阅读时间需要 9 分钟。

题目如下：

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 (<= 10⁵) 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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1

这是一道著名坑题，如果真的用单向链表来做，实在是太变态了，下面给出一种利用map和vector完成的简单方法。

这道题最大的坑在于有无效结点，也就是说除了从头结点走到-1的所有结点外，还有其他子链表，排除的方法很简单，只要从头到尾走一遍记录下来，其他的全部扔掉。

要解决这个问题，需要考虑以下几个方面：

①如何根据题目输入存储结点？

采用结构体数组，规模为10万，下标即为自己的地址，结构体内存储编号和下一个地址。

②如何进行反转？

只用vector记录有效结点的编号，同时在记录时用map记录编号到地址的对应关系。

只反转vector中的编号。

③如何连接地址输出？

通过反转后的vector，结合map查询地址，实现地址连接。

一定要注意处理K=N和N=1的情况，注意地址的前导0，-1不能有前导0。

代码如下：

#include 
    
     #include 
     
      #include 
      #include 
       
        using namespace std;struct Node{    int me;    int num;    int next;}nodes[100000];int main(){    int N;    map
        
          addMap;    int add,num,next;    int head,K;    cin >> head >> N >> K;    for(int i = 0; i < N; i++){        scanf("%d%d%d",&add,&num,&next);        nodes[add].me = add;        nodes[add].num = num;        nodes[add].next = next;        addMap[num] = add;    }    vector
         
           validList; add = head; while(add != -1){ Node n = nodes[add]; validList.push_back(n.num); add = n.next; } vector
          
            reverseList; int len = validList.size(); int cur = len; for(int group = 0; group * K < len; group++){ cur = group * K; if(len - cur < K) break; else{ cur = len; for(int i = K - 1; i >= 0; i--){ reverseList.push_back(validList[group * K + i]); } } } for(int i = cur; i < len; i++){ reverseList.push_back(validList[i]); } len = reverseList.size(); if(len == 1){ int num = reverseList[0]; printf("%05d %d -1\n",addMap[num],num); }else{ for(int i = 0; i < len - 1; i++){ int now = reverseList[i]; int next = reverseList[i+1]; printf("%05d %d %05d\n",addMap[now],now,addMap[next]); } int now = reverseList[len - 1]; printf("%05d %d -1\n",addMap[now],now); } return 0;}

转载于:https://www.cnblogs.com/aiwz/p/6154052.html

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