PAT (Advanced) 1074. Reversing Linked List (25)

根据评论里的那种思路写的一个代码。这种思路很简单，用一个数组保存每个节点的地址，然后对数组进行reverse。这个题目还可以运用map，也非常方便。

#include<iostream>
#include<cstring>
#include <algorithm>
using namespace std;


int node[100002][2];

int main()
{
	int list[100002];
	int i;
	int st, k;
	int n;
	cin >> st >> n >> k;
	for (i = 0; i<n; i++)
	{
		int add, data, next;
		cin >> add >> data >> next;
		node[add][0] = data;
		node[add][1] = next;
	}
	int cnt = 0;
	int cur = st;
	while (cur != -1)
	{
		list[cnt++] = cur;
		cur = node[cur][1];
	}
	
	i = 0;
	while (i + k <= cnt)
	{
		reverse(list + i, list + i + k);
		i += k;
	}
	for (i = 0; i < cnt - 1; i++)
	{
		printf("%05d %d %05d\n", list[i], node[list[i]][0], list[i+1]);
	}
	printf("%05d %d -1\n", list[i], node[list[i]][0]);
	return 0;
}

之前写了一个代码，在最底下，有bug，最后一个1分的case无法通过，原因是没有考虑到输入的N个节点，有某个或某些不在链表上，所以最后要输出的总的节点数小于N。

这个题目我用的方法非常蛋疼，写了很久。这个题目还有很多取巧的办法。我完整通过的代码如下：

#include <iostream>
#include <vector>

using namespace std;

struct Node
{
	int data, next;
}node[100005];

void printList(int start)
{
	int cur = start;
	while (node[cur].next != -1)
	{
		printf("%05d %d %05d\n", cur, node[cur].data, node[cur].next);
		cur = node[cur].next;
	}
	if (node[cur].next == -1)
	{
		printf("%05d %d -1\n", cur, node[cur].data);
	}
}

int reverseK(int start, int n, int k)
{
	int head, pre, cur, next, pos, len1, len2, times = n / k;
	vector<int> first, last;
	first.push_back(start);

	for (int i = 0; i < times; i++)
	{
		pre = -1;
		cur = start;
		for (pos = 1; pos <= k; pos++)
		{
			next = node[cur].next;
			if (pos == k)
			{
				if (i == 0)
					head = cur;
				last.push_back(cur);
				first.push_back(next);
			}
			node[cur].next = pre;
			pre = cur;
			cur = next;
		}
		start = next;
	}
	len1 = first.size(), len2 = last.size();
	for (int i = 0; i < len2 - 1; i++)
		node[first[i]].next = last[i + 1];
	if (len1 > len2)
		node[first[len1 - 2]].next = first[len1 - 1];
	
	return head;
}

int main()
{
	int head, N, K;
	cin >> head >> N >> K;
	int len = 1;
	int Address, Data, Next;
	for (int i = 0; i < N; i++)
	{
		cin >> Address >> Data >> Next;
		node[Address].data = Data;
		node[Address].next = Next;
	}
	int cur = head;
	while (node[cur].next != -1)
	{
		len++;
		cur = node[cur].next;
	}
	if (K != 1)
		head = reverseK(head, len, K);
	printList(head);
	return 0;
}

before：

方法非常蛋疼，最后一个1分的case始终过不了，蛋疼。代码如下，等待改进。。。

ps:如果你发现了我代码中的bug，为什么过不了，以及如果你有更好的方法，请一定要留言告诉我，非常感谢！

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <vector>

using namespace std;

struct Node
{
	int data, next;
}node[100005];

void printList(int start, int n)
{
	int cur = start;
	for (int i = 0; i < n; i++)
	{
		printf((node[cur].next == -1 ? "%05d %d %d\n" : "%05d %d %05d\n"), cur, node[cur].data, node[cur].next);
		cur = node[cur].next;
	}
}

int reverseK(int start, int n, int k)
{
	int head, pre, cur, next, pos, len1, len2, times = n / k;
	vector<int> first, last;
	first.push_back(start);

	for (int i = 0; i < times; i++)
	{
		pre = -1;
		cur = start;
		for (pos = 1; pos <= k; pos++)
		{
			next = node[cur].next;
			if (pos == k)
			{
				if (i == 0)
					head = cur;
				last.push_back(cur);
				first.push_back(next);
			}
			node[cur].next = pre;
			pre = cur;
			cur = next;
		}
		start = next;
	}
	len1 = first.size(), len2 = last.size();
	for (int i = 0; i < len2 - 1; i++)
		node[first[i]].next = last[i + 1];
	if (len1 > len2)
		node[first[len1 - 2]].next = first[len1 - 1];
	
	return head;
}

int main()
{
//	freopen("in.txt", "r", stdin);
	int head, N, K;
	cin >> head >> N >> K;

	int Address, Data, Next;
	for (int i = 0; i < N; i++)
	{
		cin >> Address >> Data >> Next;
		node[Address].data = Data;
		node[Address].next = Next;
	}
	if (K != 1)
		head = reverseK(head, N, K);
	printList(head, N);
	return 0;
}