(PAT 1097) Deduplication on a Linked List （链表)

最新推荐文章于 2024-04-22 15:17:43 发布

alex1997222

最新推荐文章于 2024-04-22 15:17:43 发布

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分类专栏： ACM算法习题

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本文介绍了一种算法，用于处理单链表中绝对值重复的整数键值，仅保留每个绝对值首次出现的节点，并将移除的节点收集到另一个链表中。通过使用哈希表记录元素绝对值的出现次数，该算法实现了高效地遍历和修改链表。

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Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, given L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (≤105) which is the total number of nodes. 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 Key Next

where Address is the position of the node, Key is an integer of which absolute value is no more than 104, and Next is the position of the next node.

Output Specification:

For each case, output the resulting linked list first, then the removed list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 5
99999 -7 87654
23854 -15 00000
87654 15 -1
00000 -15 99999
00100 21 23854

Sample Output:

00100 21 23854
23854 -15 99999
99999 -7 -1
00000 -15 87654
87654 15 -1

解题思路:利用哈希表保存各个元素绝对值的出现次数

所有结点都设置一个前置结点

删除结点点时,使这个结点的前置结的后置结点指向删除结点的后置结点,删除结点后置结点的前置结点指向删除结点前置,这个删除结点就被删除了

创建新链表时,每次保存当前结点地址作为下一个结点的前置即可

#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
const int MAXN = 100010;
const int MAXNUM = 10010;
struct Node {
	int data;
	int preAddr;
	int Addr;
	int nextAddr;
}nodes[MAXN],Dnodes[MAXN];
int hashTable[MAXNUM] = { 0 };
int main() {
	int sAddr, N;
	scanf("%d %d", &sAddr, &N);
	for (int i = 0; i < N; ++i) {
		int Addr, nextAddr, data;
		scanf("%d %d %d", &Addr, &data, &nextAddr);
		nodes[Addr].data = data;
		nodes[Addr].nextAddr = nextAddr;
		nodes[Addr].Addr = Addr;
	}
	//链表遍历,建立前向关系
	int p1 = sAddr;
	int preAddr = -1;
	while (p1 != -1) {
		nodes[p1].preAddr = preAddr;
		preAddr = p1;
		p1 = nodes[p1].nextAddr;
	}
	p1 = sAddr;
	int tempPreAddr = -1;
	int tempHead = -1;
	while (p1 != -1) {
		hashTable[abs(nodes[p1].data)]++;
		if (hashTable[abs(nodes[p1].data)] > 1) {  //出现重复
			nodes[nodes[p1].preAddr].nextAddr = nodes[p1].nextAddr;
			if (nodes[p1].nextAddr != -1) {
				nodes[nodes[p1].nextAddr].preAddr = nodes[p1].preAddr;
			}
			//加入到删除链表
			//删除链表为空的情况
			if (tempPreAddr == -1 && tempHead == -1) {
				Dnodes[p1].data = nodes[p1].data;
				Dnodes[p1].nextAddr = tempPreAddr;
				tempPreAddr = p1;
				tempHead = p1;
			}
			else {  //不为空的情况
				Dnodes[tempPreAddr].nextAddr = p1;
				Dnodes[p1].data = nodes[p1].data;
				Dnodes[p1].nextAddr = -1;
				tempPreAddr = p1;
			}
		}
		p1 = nodes[p1].nextAddr;
	}
	p1 = sAddr;
	while (p1 != -1) {
		if (nodes[p1].nextAddr == -1) {
			printf("%05d %d %d\n", p1, nodes[p1].data, nodes[p1].nextAddr);
		}
		else {
			printf("%05d %d %05d\n", p1, nodes[p1].data, nodes[p1].nextAddr);
		}
		p1 = nodes[p1].nextAddr;
	}
	while(tempHead != -1){
		if (Dnodes[tempHead].nextAddr == -1) {
			printf("%05d %d %d\n", tempHead, Dnodes[tempHead].data, Dnodes[tempHead].nextAddr);
		}
		else {
			printf("%05d %d %05d\n", tempHead, Dnodes[tempHead].data, Dnodes[tempHead].nextAddr);
		}
		tempHead = Dnodes[tempHead].nextAddr;
	}
	system("PAUSE");
	return 0;
}