1074 Reversing Linked List

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

题意：

　　给出一个用数字表示指针的链表，将链表中的每k个数字进行反转。

思路：

　　本来想的是用结构体来存储，可是提交的时候，有一个测试点超时，还有一个测试点没有的通过。

　　date[address]表示链表所在结点中的值，next[address]表示链表所指向的下一个结点，list[]用来存储排序好的数组。有一点需要注意就是并不是所有的结点都会出现在结果中。有一个地方挺巧妙的，就是再进行反转的时候下标的确定方法：i / k * k + k - 1 - (i % k).

Code:

#include <bits/stdc++.h>

using namespace std;

int main() {
    int start, n, k;
    cin >> start >> n >> k;
    int address, val, next;
    int date[100005], nextNode[100005], list[100005], ans[100005];
    for (int i = 0; i < n; ++i) {
        cin >> address;
        cin >> date[address] >> nextNode[address];
    }
    int first = start, index = 0;
    while (first != -1) {
        list[index++] = first;
        first = nextNode[first];
    }
    for (int i = 0; i < index; ++i) ans[i] = list[i];
    for (int i = 0; i < index - index % k; ++i)
        ans[i] = list[i / k * k + k - i % k - 1];
    for (int i = 0; i < index - 1; ++i) {
        printf("%05d %d %05d
", ans[i], date[ans[i]], ans[i + 1]);
    }
    printf("%05d %d -1
", ans[index - 1], date[ans[index - 1]]);

    return 0;
}