}intcount =1;while(start!=-1){if(hasabs.find(abs(node[start].key))==hasabs.end()){if(count!=1)printf("%05d\n",start);//保留5位hasabs.insert(abs(node[start].key));printf("%05d %d ",start,node[start].key); }else{ sepadd.push_back(node[start].address); } start = node[...
(PAT 1097) Deduplication on a Linked List (链表) 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...
PAT 甲级 1097 Deduplication on a Linked ListAAA专业手机贴膜小张 立即播放 打开App,流畅又高清100+个相关视频 更多187 -- 9:24 App PAT 甲级 1074 Reversing Linked List 40 -- 8:43 App PAT 甲级 1133 Splitting A Linked List 3296 6 13:26 App pta 7-3判断上三角矩阵 142 1 13:25 App pat...
for each valueK, only the first node of which the value or absolute value of its key equalsK will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, givenL being 21→-15→-15→-7...
PAT_甲级_1097 Deduplication on a Linked List (25point(s)) (C++)【删除链表重复节点】,目录1,题目描述题目大意2,思路3,AC代码4,解题过程第一搏第二搏1,题目描述SampleInptput:0010021238542385...
PAT_甲级_1097 Deduplication on a Linked List 题目大意: 给定一条单链表,要求分解为两条链表,第一条是去除绝对值重复数值结点的链表,第二条是去除的结点组成的链表 算法思路: 首先使用node存储所有输入的节点,removed存储需要删除的节点。由于需要判断node中的节点是否需要删除,我们使用isExist表示与该节点数据的...
step2 柳神的代码,思路是差不多的,直接上。 #include<cstdio>#include<stdlib.h>#include<algorithm>usingnamespacestd;constintmaxn=100000;structNODE{intaddress,key,next,num=2*maxn;}node[maxn];boolexist[maxn];intcmp1(NODEa,NODEb){returna.num<b.num;}intmain(){intbegin,n,cnt1=0,cnt2=0,...
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...
#include<iostream>#include<vector>usingnamespacestd;structnode{intkey,next;}n[100000];intmain(){inti,j,k;inthead,N;cin>>head>>N;while(N--){cin>>i>>j>>k;//i地址;j数据;k为nextn[i].key=j;n[i].next=k;}//输入数据boolappeared[100000]={0};vector<int>v1,v2;for(i=head;i>...
//思路:第一步用node将链表链接起来 存在v中 用ma判断是否重复 重复则pop 并push到re中#include #include #include #include #include using na...