PAT甲级 链表题_C++题解

链表处理

PAT (Advanced Level) Practice 链表题

目录

  • 《算法笔记》 重点摘要：静态链表
  • 1032 Sharing (25)
  • 1052 Linked List Sorting (25)
  • 1097 Deduplication on a Linked List (25)
  • 1133 Splitting A Linked List (25)
  • 附： 动态链表

《算法笔记》 7.3 链表 重点摘要

静态链表 ⭐

(1) 定义

结构体类型名和结构体变量名尽量不同

struct Node{
    int address; // 记录地址（若后面排序则不能以数组下标代替）
    typename data;
    int next;
    XXX...
} node[100001];
(2) 初始化
  • 遍历所有结点，将性质变量标记为比正常情况（如在所给链表上）下更小的数字
for (int i = 0; i < 100001; i++){
    node[i].XXX = 0;
}
  • 输入节点时直接将性质变量的初始值压入
int n, head, address, key, next;
scanf("%d%d", &n, &head);
for (int i = 0; i < n; i++){
	scanf("%d%d%d", &address, &key, &next);
	node[address] = {address, key, next, 0};
}
(3) 标记

依据题目所给链表首地址和各结点信息遍历链表上的结点，标记性质，记录个数

int p = head, count = 0;
while (p != -1){
    XXX = 1;
    count++;
    p = node[p]->next;
}

int count = 0;
for (int p = head; p != -1; p = node[p].next){
    node[p].XXX = 1;
    count++;
}
(4) 排序
bool cmp (Node a, Node b){
    if (a.XXX == -1 || b.XXX = -1){
        // 比较标记性质，若有无效点此性质较小 -> 被放到数组后面
        return a.XXX > b.XXX;
    }
    else{
        // 按要求排序
    }
}

1032 Sharing (25)

题目思路

  • 链表结点中添加性质变量 inFirst 记录是否在第一个链表中
  • 初始化时将 inFirst 均置为 false
  • 结点输入完毕后，遍历第一条链表，令结点 inFirst = true
  • 遍历第二条链表，检查结点 inFirst，若为 true，break跳出循环
  • 检查第二条链表循环变量
    • 若未到链表尾，说明找到了和第一条链表共享的结点，按 %05d 输出地址
    • 若为 -1 说明到链表尾未找到，输出 -1
#include<iostream>
using namespace std;
struct Node{
	char data;
	int next;
	bool inFirst;
} node[100001];
int main()
{
	int head1, head2, n, address, next, p;
	char data;
	scanf("%d%d%d", &head1, &head2, &n);
	for (int i = 0; i < n; i++){
		scanf("%d %c %d", &address, &data, &next);
		node[address] = {data, next, false};
	}
	for (p = head1; p != -1; p = node[p].next)
		node[p].inFirst = true;
	for (p = head2; p != -1; p = node[p].next)
		if (node[p].inFirst == true) break;
	if (p == -1) printf("-1");
	else printf("%05d", p);
	return 0;
}

1052 Linked List Sorting (25)

题目思路

  • 结点中添加性质变量 inList，用 int 型的 0/1 来表示，便于排序中比较
  • 结点中要保留此结点的地址，因为按 key 排序后数组下标将不再能表示地址
  • 输入完成后，从头结点遍历链表，令 inList = 1 并 记录链表中的结点数
  • 排序函数将不在链表中的放到后面，在链表中的按 key 值排序，故排序后数组的前 count 个即为新链表
  • 输出时注意数组前一个结点的新 next 应当为数组下一个结点的地址，而非原链表的 next
#include<iostream>
#include<algorithm>
using namespace std;
struct Node{
	int address, key, next, inList;
} node[100001];
bool cmp (Node a, Node b){ return !a.inList || !b.inList ? a.inList > b.inList : a.key < b.key; }
int main()
{
	int n, head, address, key, next, count = 0;
	scanf("%d%d", &n, &head);
	for (int i = 0; i < n; i++){
		scanf("%d%d%d", &address, &key, &next);
		node[address] = {address, key, next, 0};
	}
	for (int i = head; i != -1; i = node[i].next){
		node[i].inList = 1;
		count++;
	}
	if (!count) printf("0 -1\n");
	else{
		sort(node, node+100001, cmp);
		printf("%d %05d\n", count, node[0].address);
		for (int i = 0; i < count - 1; i++) printf("%05d %d %05d\n", node[i].address, node[i].key, node[i+1].address);
		printf("%05d %d %d\n", node[count-1].address, node[count-1].key, -1);
	}
	return 0;
}

1097 Deduplication on a Linked List (25)

题目思路

  • 读入静态链表结点数据，用一个 set 记录出现过的数字的绝对值
  • 扫描题目给出的链表结点
    • 若其数据域绝对值已经出现过，则放到 removed 容器中
    • 若未出现过，将数据域绝对值加入到集合中，结点放到 result 容器中
  • 先输出 result 容器，再输出 removed 容器
  • 注意！！！ 容器的大小size()返回值为一个无符号数，无符号数运算后再比较十分危险
    • for (int i = 0; i < vector.size() - 1; i++ 改为 for (int i = 1; i < vector.size(); i++ 避开 size()-1
#include<iostream>
#include<cmath>
#include<set>
#include<vector>
using namespace std;
struct Node{
	int address, data, next;
} node[100001];
int main()
{
	int head, n, address, data, next;
	scanf("%d%d", &head, &n);
	for (int i = 0; i < n; i++){
		scanf("%d%d%d", &address, &data, &next);
		node[address] = {address, data, next};
	}
	vector<Node> result, removed;
	set<int> occured;
	for (int p = head; p != -1; p = node[p].next){
		if (occured.find(abs(node[p].data)) != occured.end()) removed.push_back(node[p]);
		else{
			occured.insert(abs(node[p].data));
			result.push_back(node[p]);
		}
	}
	for (int i = 1; i < result.size(); i++)
		printf("%05d %d %05d\n", result[i-1].address, result[i-1].data, result[i].address);
	if (result.size() > 0) printf("%05d %d -1\n", result[result.size()-1].address, result[result.size()-1].data);
	for (int i = 1; i < removed.size(); i++)
		printf("%05d %d %05d\n", removed[i-1].address, removed[i-1].data, removed[i].address);
	if (removed.size() > 0) printf("%05d %d -1\n", removed[removed.size()-1].address, removed[removed.size()-1].data);
	return 0;
}

1133 Splitting A Linked List (25)

题目思路

  • 设置全局变量 K，在程序中接收输入，以便于 cmp 函数使用
  • 在结点中设置 oriorder 变量，记录其在原始链表中的顺序，以便于 cmp 函数使用
  • cmp 函数首先将不在链表中的放到数组后面，再看如果两个结点数值分别在 0 或 K 两侧，则按数值域排序，否则仍按原链表顺序
#include<iostream>
#include<algorithm>
using namespace std;
int K = 0;
struct Node{
	int address, data, next, inList, oriorder;
} node[100001];
bool cmp (Node a, Node b){
	if (!a.inList || !b.inList) return a.inList > b.inList;
	if ((a.data >= 0 && b.data < 0) || (a.data < 0 && b.data >= 0)) return a.data < b.data;
	if ((a.data > K && b.data <= K) || (a.data <= K && b.data > K)) return a.data < b.data;
	return a.oriorder < b.oriorder;
}
int main()
{
	int n, head, address, data, next, count = 0, order = 0;
	scanf("%d%d%d", &head, &n, &K);
	for (int i = 0; i < n; i++){
		scanf("%d%d%d", &address, &data, &next);
		node[address] = {address, data, next, 0, 0}; 
	}
	for (int p = head; p != -1; p = node[p].next){
		node[p].inList = 1;
		node[p].oriorder = order;
		count++;
		order++;
	}
	sort(node, node+100001, cmp);
	for (int i = 0; i < count - 1; i++)
		printf("%05d %d %05d\n", node[i].address, node[i].data, node[i+1].address);
	printf("%05d %d %d\n", node[count-1].address, node[count-1].data, -1);
	return 0;
}

二刷（参考柳婼小姐姐的代码）

  • 遍历链表，先后选取 (-无穷,0)， [0,k], (k,+无穷)压入 result 容器
  • 输出 result 中的结点即为所要求的顺序
#include<iostream>
#include<vector>
using namespace std;
struct Node{
	int address, data, next;
} node[100000];
int main()
{
	int head, n, k, address, data, next;
	scanf("%d%d%d", &head, &n, &k);
	for (int i = 0; i < n; i++){
		scanf("%d%d%d", &address, &data, &next);
		node[address] = {address, data, next};
	}
	vector<Node> result;
	for (int p = head; p != -1; p = node[p].next)
		if (node[p].data < 0) result.push_back(node[p]);
	for (int p = head; p != -1; p = node[p].next)
		if (node[p].data >= 0 && node[p].data <= k) result.push_back(node[p]);
	for (int p = head; p != -1; p = node[p].next)
		if (node[p].data > k) result.push_back(node[p]);
	for (int i = 1; i < result.size(); i++)
		printf("%05d %d %05d\n", result[i-1].address, result[i-1].data, result[i].address);
	printf("%05d %d -1\n", result[result.size()-1].address, result[result.size()-1].data);
	return 0;
}

附：

动态链表

  • 头结点：数据域不存放任何内容，指针域指向第一个数据域有内容的结点
  • 最后一个节点的指针域指向 NULL，表示链表结尾
(1) 创建链表
node *p, *pre, *head;
head  = new node();
head->next = NULL;
pre = head;
for (int i = 0; i < n; i++){
    p = new node;
    p->data = data[i];
    p->next = NULL;
    pre->next = p;
    pre = p;
}
(2) 查找元素
// 查找链表中 x 并记录个数
node *p = head->next;
while (p != NULL){
    if (p->data == x){
        count++;
    }
    p = p->next;
}
(3) 插入元素
// 将 x 插入链表第 pos 个位置
node *p = head;
for (int i = 0; i < pos-1; i++){
    p = p->next;
}
node *q = new node;
q->data = x;
q->next = p->next;
p->next = q;
(4) 删除元素
// 删除链表中数据域为 x 的结点
node *p = head->next;
node *pre = head;
while (p != NULL){
    if (p->data = x){
        pre->next = p->next;
        delete(p);
        p = pre->next;
    }
    else{
        pre = p;
        p = p->next;
    }
}
posted @ 2019-09-06 13:26  鲸90830  阅读(469)  评论(0)  编辑  收藏  举报