PTA_2019春_023_Reversing Linked List

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最新推荐文章于 2022-11-27 18:22:56 发布

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分类专栏：数据结构PTA

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本文介绍了一种链表操作算法，即给定一个链表和一个常数K，将链表上的每K个元素进行反转。通过将链表分割成小链表，分别反转并重新连接，实现了链表的K个一反转。特别讨论了K为1和K等于链表长度的情况。

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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

Sample Output`:`

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

简述：将给定链表按所给K值，每K个反转，最后不满足K个的不反转。

思路：1）将链表按每K个分割成小链表；

2）每个小链表反转；

3）将小链表连接起来，并将剩余的接上。

特别考虑：当K=1时，链表不变。K=N时，整体反转。

输入的节点中可能有部分不在链表上。

反思：将小链表连接起来时存在逻辑错误，导致链表并没有连起来。

/*此题本不难
  奈何愁三载 
  缘由我很菜
  前进向前看*/
#include<stdio.h>
#define MAXSIZE 100001
struct LNode {
	int Data;
	int Next;
};
int Reverse(struct LNode* Array, int K ,int Firstaddress ,int Number);
void PRT(struct LNode* Array, int Firstaddress);
int TrueNumber(struct LNode* Array, int Firstaddress);
int main() {
	/*将输入存入结构体数组中*/
	int Firstaddress, Number, K;
	int Address, Data, Next;
	scanf("%d %d %d", &Firstaddress, &Number, &K);
	struct LNode Array[MAXSIZE];
	for (int i = 0; i < Number; i++) {
		scanf("%d %d %d", &Address, &Data, &Next);
		Array[Address].Data = Data;
		Array[Address].Next = Next;
	}
	Number =  TrueNumber(Array,Firstaddress);/*除去多余的节点，返回链表真正的个数*/
	if(K<=Number&&K>1){
		Firstaddress = Reverse(Array, K, Firstaddress, Number);
	}
	PRT(Array, Firstaddress);
	return 0;
}
int Reverse(struct LNode* Array, int K,int Firstaddress ,int Number) {
	int Old_head, New_head, Temp;/*反转时需要*/ 
	int last_first;/*记录上一个循环最后一个点*/
	int last_first_label;/*第一次循环不需要*/ 
	Old_head = Firstaddress;
	New_head = -1;
	last_first_label = -1;
	for (int i = 0; i < Number / K; i++) {
		int j = 0;
		New_head = -1;
		/*以下实现反转*/
		while (j < K&&Old_head!=-1) {
			if (j == 0) last_first = Old_head;/*标记这个循环下的最后一个节点*/
			Temp = Array[Old_head].Next;
			Array[Old_head].Next = New_head;
			New_head = Old_head;
			Old_head = Temp;
			j++;
		}
		if (i == 0) Firstaddress = New_head;/*首地址*/
		if (last_first_label != -1) {
			Array[last_first_label].Next = New_head;
		}
		last_first_label=last_first;
	}/*结束时Old_head指向下一个节点*/
	if (Old_head == -1) return Firstaddress;/*此时节点个数正好被K整除*/
	else {/*将剩下节点直接接在顺序之后*/
		Array[last_first].Next = Old_head;
		return Firstaddress;
	}
}
void PRT(struct LNode* Array, int Firstaddress) {
	int address;
	address = Firstaddress;
	while (Array[address].Next!=-1) {
		printf("%05d %d %05d\n",address,Array[address].Data,Array[address].Next);
		address = Array[address].Next;
	}
	printf("%05d %d %d", address, Array[address].Data, Array[address].Next);
}
int TrueNumber(struct LNode* Array, int Firstaddress){
	int cnt = 0;
	int address;
	address = Firstaddress;
	while(address!=-1){
		address=Array[address].Next;
		cnt++;
	}
	return cnt;
}