Advanced Level 1074 Reversing Linked List (25 point(s))

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 ( ≤ 1 0 5 ) N (≤10^5) 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

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

思路分析:

还应该考虑输出中有不再链表的结点,所以用sum计数。可以调用algorithm中的reverse函数。

代码:

#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100010;
int head, idx, e[maxn], ne[maxn], n, k, list[maxn];
int main(){
    cin >> head >> n >> k;
    for(int i = 0; i < n; i++){
        cin >> idx;
        cin >> e[idx] >> ne[idx];
    }
    int sum = 0;
    while(head != -1){
        list[sum++] = head;
        head = ne[head];
    }
    for(int i = 0; i < sum / k; i += k)
        reverse(list + i, list + i + k);
    for(int i = 0; i < sum - 1; i++)
        printf("%05d %d %05d\n", list[i], e[list[i]], list[i + 1]);
    printf("%05d %d -1", list[sum - 1], e[list[sum - 1]]);
    return 0;
}

PAT_Advanced_Level

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