zoukankan      html  css  js  c++  java
  • 1052. Linked List Sorting (25)

    A linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and a Next pointer to the next structure. Now given a linked list, you are supposed to sort the structures according to their key values in increasing order.

    Input Specification:

    Each input file contains one test case. For each case, the first line contains a positive N (< 105) and an address of the head node, where N is the total number of nodes in memory and the address of a node is a 5-digit positive integer. NULL is represented by -1.

    Then N lines follow, each describes a node in the format:

    Address Key Next

    where Address is the address of the node in memory, Key is an integer in [-105, 105], and Next is the address of the next node. It is guaranteed that all the keys are distinct and there is no cycle in the linked list starting from the head node.

    Output Specification:

    For each test case, the output format is the same as that of the input, where N is the total number of nodes in the list and all the nodes must be sorted order.

    Sample Input:

    5 00001
    11111 100 -1
    00001 0 22222
    33333 100000 11111
    12345 -1 33333
    22222 1000 12345
    

    Sample Output:

    5 12345
    12345 -1 00001
    00001 0 11111
    11111 100 22222
    22222 1000 33333
    33333 100000 -1

    解题过程:比较坑,没考虑head=-1的情况,最后一个测试用例报内存超时。
    #include<iostream>
    #include<cstdio>
    #include<string>
    #include<vector>
    #include<algorithm>
    using namespace std;
    struct Node{
    	int add;
    	int key;
    	int next;
    }; 
    vector<Node>v;
    int index[100002];
    int val[100002];
    bool cmp(Node a,Node b){
    	return a.key<b.key;
    }
    int main(){
    	int i,j,n;
    	int head;
    	scanf("%d%d",&n,&head);
    	int add,key,next;
    	Node node;
    	for(i=0;i<n;i++){
    		scanf("%d%d%d",&add,&key,&next);
    		index[add]=next;
    		val[add]=key;
    	}
    	add=head;
    	if(add == -1){
    		printf("0 -1
    ");
    		return 0;
    	}
    	while(1){
    		node.add=add;
    		node.key=val[add];
    		node.next=index[add];
    		v.push_back(node);
    		add=index[add];
    		if(add==-1){
    			break;
    		}
    	}
    	sort(v.begin(),v.end(),cmp);
    	int size=v.size();
    	head=v[0].add;
    	for(i=0;i<size-1;i++){
    		v[i].next=v[i+1].add;
    	} 
    	v[size-1].next=-1;
    	
    	printf("%d %05d
    ",size,head);
    	for(i=0;i<size-1;i++){
    		printf("%05d %d %05d
    ",v[i].add,v[i].key,v[i].next);
    	}
    	printf("%05d %d -1
    ",v[i].add,v[i].key);
    	return 0;
    }
    

      




  • 相关阅读:
    如何避免JavaScript的内存泄露及内存管理技巧
    【跟我一起学python吧】python chr()、unichr()和ord()
    阿里云,实力与担当并存!
    首届阿里白帽大会成功举办,用技术“肩天下”
    DataV数据可视化年度峰会——唤醒数据,看见未来
    支付宝移动端 Hybrid 解决方案探索与实践
    大数据上云第一课:(1)MaxCompute授权和外表操作躲坑指南
    函数计算支持应用中心功能
    Serverless 解惑——函数计算如何访问 MySQL 数据库
    开发函数计算的正确姿势——使用交互模式安装依赖
  • 原文地址:https://www.cnblogs.com/grglym/p/7786666.html
Copyright © 2011-2022 走看看