02-线性结构3 Reversing Linked List

题目描述

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 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
结尾无空行

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1
结尾无空行

代码实现

#include <stdio.h>

#define MAX_SIZE 100000

typedef struct Node
{
    int data;
    int next;
} MyNode;

// [first, right)区间对应的节点翻转
void MyReverse(int *list, int first, int last)
{
    while (first < last) {
        int tmp = list[first];
        list[first] = list[last];
        list[last] = tmp;
        first++;
        last--;
    }
}

int main()
{
    int firstAddr;
    int n;
    int k;
    scanf("%d%d%d", &firstAddr, &n, &k);

    int addr;
    int data;
    int next;
    
    // 数组小标表示addr
    MyNode node[MAX_SIZE];

    for (int i = 0; i < n; i++) {
        scanf("%d%d%d", &addr, &data, &next);
        node[addr].data = data;
        node[addr].next = next;
    }
    
    int end = 0;
    int cur = firstAddr;
    
    int addrList[MAX_SIZE];
    while(cur != -1)
    {
        addrList[end++] = cur;
        cur = node[cur].next;
    }

    /*
    for (int i = 0; i < end; i++) {
        printf("%d %d\n", i, addrList[i]);
    }
    */
    
    int first = 0;
    while (first + k <= end)
    {
        MyReverse(addrList, first, first + k - 1);
        first += k;
    }

    for(first = 0; first < end - 1; first++)
        printf("%05d %d %05d\n", addrList[first], node[addrList[first]].data, addrList[first + 1]);
    printf("%05d %d -1\n", addrList[first], node[addrList[first]].data);
    
    return 0;
}