PAT (Advanced Level) Practice 1074 Reversing Linked List（25分）【链表】

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

题意

分组反转链表。

思路

不是所有的节点都是有效节点，首先把链表上的节点按照顺序读出来，便于后面的倒置操作。

代码

#include <algorithm>
#include <iostream>

using namespace std;

struct node {
    int address, data, next;
} nodes[100005], list[100005];

int main() {
    int head, n, k, addr;
    cin >> head >> n >> k;

    for (int i = 0; i < n; ++i) {
        cin >> addr;
        nodes[addr].address = addr;
        cin >> nodes[addr].data >> nodes[addr].next;
    }

    int len = 0;
    for (int now = head; now != -1; now = nodes[now].next)  // 获得链表
        list[len++] = nodes[now];
    for (int i = k; i <= len; i += k)  // 分组倒置
        reverse(list + i - k, list + i);
    for (int i = 0; i < len - 1; ++i)
        printf("%05d %d %05d\n", list[i].address, list[i].data, list[i + 1].address);
    printf("%05d %d -1\n", list[len - 1].address, list[len - 1].data);
}