1074 Reversing Linked List (25分)

标签：复制   -o   font   link   ted   第一个   ever   sed   ecif   

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

题目大意：1、first(第一个元素的地址) num(元素的个数) k(每k个元素进行翻转)

　　　　　2、输入num个“Address Data Next”类型的元素，以first为首元结点地址，Next为下一个结点地址。生成一个以“-1”为尾元结点的Next的链表后，对链表的每k个元素进行翻转。

 1 #include <iostream>
 2 using namespace std;
 3 int main() {
 4     int first, num, k, sum = 0;
 5     cin >> first >> num >> k;
 6     int temp, data[100000], next[100000], addr1[100000], addr2[100000];
 7     for (int i = 0; i < num; i++) {
 8         cin >> temp;
 9         cin >> data[temp] >> next[temp];
10     }
11     while (first != -1) {
12         addr1[sum++] = first;
13         first = next[first];
14     }
15     for (int i = 0; i < sum; i++)  addr2[i] = addr1[i];
16     for (int i = 0; i < (sum - sum % k); i++)  addr2[i] = addr1[i / k * k + k - (1 + i % k)];
17     for (int i = 0; i < sum - 1; i++)
18         printf("%05d %d %05d\n", addr2[i], data[addr2[i]], addr2[i + 1]);
19     printf("%05d %d -1", addr2[sum - 1], data[addr2[sum - 1]]);
20     return 0;
21 }

解法： 1、用临时变量temp作为数组指针，把地址为temp的元素数值存入data[temp]中，把temp的元素下一个结点的地址存入next[temp]中,用sum记录链表元素个数。
　　　 2、通过first变量用迭代器的功能实现满足链表顺序的addr1数组
　　　 3、addr1数组经过每k个元素进行翻转变成addr2；
　　　　　　*先把addr1数组复制成addr2数组；
　　　　　　*再对(sum-sum%k)（即满足翻转要求的前x组元素）个元素进行翻转，其中：1/k*k（欲翻转元素i前面的整组元素个数，也就是该组的第一个地址）、k - (1 + i % k)（每组翻转前i应该对应加的几个地址）

1074 Reversing Linked List (25分)

标签：复制   -o   font   link   ted   第一个   ever   sed   ecif   

原文地址：https://www.cnblogs.com/i-chase/p/13154997.html