PAT-A 1074 Reversing Linked List （25 分)

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 Lbeing 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

Sample Output:

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

思路：给出的节点可能不在链表中，需要特别注意。这个解法复杂了，和一年前的思路完全不同。更简练的代码参考B1025的解法。https://mp.csdn.net/postedit/81151398

#include <queue>
#include <map>
#include <stack>using namespace std;typedef struct node
{int addr;int data;int next;node(){}node(int a, int d, int n):addr(a), data(d), next(n){}
}node;int main()
{int head, n, k;scanf("%d%d%d", &head, &n, &k);map<int, node> all;for(int i = 0; i < n; i++){int addr, data, next;scanf("%d%d%d", &addr, &data, &next);all[addr] = node(addr, data, next);}queue<node> link;int len = 0;while(head != -1){link.push(all[head]);head = all[head].next;len++;}stack<node> work;queue<node> reversed;int group = len / k; //需要逆置的组数for(int i = 0; i < group; i++){for(int j = 0; j < k; j++){work.push(link.front());link.pop();}if(!reversed.empty())reversed.back().next = work.top().addr;while(!work.empty()){reversed.push(work.top());work.pop();if(!work.empty())reversed.back().next = work.top().addr;}}while(!link.empty()){reversed.back().next = link.front().addr;reversed.push(link.front());link.pop();}reversed.back().next = -1;while(!reversed.empty()){node tem = reversed.front();int addr = tem.addr;int data = tem.data;int next = tem.next;printf("%05d %d ", addr, data);if(next == -1)printf("-1\n");elseprintf("%05d\n", next);reversed.pop();}return 0;
}