《数据结构》02-线性结构3 Reversing Linked List

╰+攻爆jí腚メ 2022-05-04 00:28 166阅读 0赞

# 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 ^5 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

# 分析 #

题目大概意思就是给一组格式是`Address data next`的结点，按要求逆序输出  
先观察，逆序后不变的地方是前两组数据，即每个结点的 Address 和 data 是不变的，继续观察，其实当前结点的 next 是下一个结点的 Address  
我们考虑将 Address 和 data 存起来，还要体现它们的关系，可以用数组 Data\[Address\] = data，同样的，我们可以将 next 和 Address 也以这种形式存储，即 Next\[Address\] = next  
这样存储的结果是，知道 Address 我们可以通过 Data\[ \] 数组找到对应 data，可以通过 Next\[ \] 数组找到对应的 next  
这样存储后，我们再把整个链表“理顺”，给出的 first node 即为第一个 Address，Next\[first node\] 是下一个结点的 Address，再用一个数组 list\[ \] 把每次遇到的 Address 记录下来  
到此为止我们有了能根据 Address 找到其他数据的关系数组，也有顺序记录 Address 的数组，问题就从"反转链表"简化为了"反转 Address"  
再以每 K 个结点为区间，反转 list\[ \] 数组中存储的 Address，再根据 Address 和其他数据的关系，就能完整地输出链表了！

#include<stdio.h>
    #include<iostream>
    #define MaxSize 100005
    using namespace std;
    int main(){
        
    	int Data[MaxSize];
    	int Next[MaxSize];
    	int list[MaxSize];
    	int FirstAdd,N,K;
    	scanf("%d %d %d",&FirstAdd,&N,&K);
    	for(int i=0;i<N;i++){
        
    		int tmpAdd,tmpData,tmpNext;
    		scanf("%d %d %d",&tmpAdd,&tmpData,&tmpNext);
    		Data[tmpAdd] = tmpData;
    		Next[tmpAdd] = tmpNext;
    	}
    	int sum=0;   // 累计有效结点数 
    	while(FirstAdd!=-1){
           // 当尾结点为 -1 时结束 
    		list[sum++] = FirstAdd;   // 记录所有Address
    		FirstAdd = Next[FirstAdd];  // 找下一个结点 
    	}
    	for(int i=0;i<sum-sum%K;i+=K){
          // 每 K 个结点一个区间 
    		for(int j=0;j<K/2;j++){
          // 反转链表
    			int t = list[i+j];
    			list[i+j] = list[i+K-j-1];
    			list[i+K-j-1] = t; 
    		}
    	}
    	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\n",list[sum-1],Data[list[sum-1]]);
    	return 0;
    }