1052. Linked List Sorting (25)

考察链表的知识，以及排序

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1052

#include<iostream>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <string>
#include <string.h>
#include <algorithm>
using namespace std;typedef struct Node
{int curAdress, key, nextAdress;bool exist;Node(){exist = false;}bool operator < (const Node& orh) const{return key < orh.key;}
}Node;
#define MAX 1000000
int main()
{int n, root;scanf("%d%d",&n,&root);vector<Node> nodeMap(MAX);for (int i = 0; i < n; ++i){Node tmp;scanf("%d%d%d",&tmp.curAdress, &tmp.key, &tmp.nextAdress);tmp.exist = true;if(tmp.curAdress >= 0 && tmp.curAdress < MAX)nodeMap[tmp.curAdress] = tmp;}//processvector<Node> nodeList;int curID = root;while(curID >= 0 && curID < MAX && nodeMap[curID].exist){nodeList.push_back(nodeMap[curID]);curID = nodeMap[curID].nextAdress;}//outputif (nodeList.empty()){if (root == -1)printf("0 -1\n");else printf("0 %05d\n",root);}else{sort(nodeList.begin(), nodeList.end());printf("%d %05d\n", nodeList.size(), nodeList[0].curAdress);for (int i = 0; i < nodeList.size(); ++i){if(i != nodeList.size()-1)printf("%05d %d %05d\n", nodeList[i].curAdress, nodeList[i].key, nodeList[i+1].curAdress);else printf("%05d %d -1\n", nodeList[i].curAdress, nodeList[i].key);}}return 0;
}

1052. Linked List Sorting (25)相关推荐

1052 Linked List Sorting (25 分)
Sample Input: 5 00001 11111 100 -1 00001 0 22222 33333 100000 11111 12345 -1 33333 22222 1000 12345 ...
【附段错误原因，最后两个测试点】1052 Linked List Sorting (25 分)【链表类题目总结】
立志用最少的代码做最高效的表达 PAT甲级最优题解-->传送门 A linked list consists of a series of structures, which are not n ...
1052 Linked List Sorting (25 分)【难度: 一般 / 知识点: 链表】
https://pintia.cn/problem-sets/994805342720868352/problems/994805425780670464 哈希表模拟链表. #include<b ...
1052 Linked List Sorting (25分)
和 1074一样的方法,只需要管自己的结点就可以了,不需要处理next, 注意最后一个测试点,测试数据最终结果为0个结点,只需要输出 0 -1 附本人AC代码: #include<iostre ...
PAT甲级1052 Linked List Sorting：[C++题解]链表排序
文章目录题目分析题目链接题目分析题意:给定数据(里面有不构成链表的数据,若是,则跳过),是链表的构成链表.然后根据数值大小重新排序,构成新的链表. 分析:用数组模拟链表,先建立链表.遍历链表, ...
1052. Linked List Sorting
没仔细审题,走了很多弯路.题目要求:内存中有很多散列的节点,可能构成不止一个链表.其次要注意空链表的情况. // 1052. Linked List Sorting.cpp: 主项目文件.#inclu ...
1052 Linked List Sorting（排序）
1052 Linked List Sorting(排序) 思路: s t r u c t + struct+ struct+排序. 坑点: 1.答案只包含从原链表中头结点开始的结点,有点结点不在原链表 ...
PAT甲级1052：Linked List Sorting (25)
题目 A linked list consists of a series of structures, which are not necessarily adjacent in memory. W ...
1052 Linked List Sorting 分数 25
题目描述 A linked list consists of a series of structures, which are not necessarily adjacent in memory. ...

1052. Linked List Sorting (25)

1052. Linked List Sorting (25)相关推荐

最新文章

热门文章