Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 105) which is the total number of nodes, and a positive K (<=1000). 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 Data Next
where Address is the position of the node, Data is an integer in [-105, 105], and Next is the position of the next node. It is guaranteed that the list is not empty.
For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.
00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218
33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1
给一个链表和K,遍历链表后将<0的结点先输出,再将0~k区间的结点输出,最后输出>k的结点
将结点用list[10000]保存,list为node类型,node中保存结点的值value和它的next地址。list的下标就是结点的地址。将<0、0~k、>k三部分的结点地址分别保存在v[0]、v[1]、v[2]中,最后将vector中的值依次输出即可~
原文链接:https://blog.csdn.net/liuchuo/article/details/78037305
#include <bits/stdc++.h> using namespace std; const int maxn=100000; struct Node { int address,key,next; }node[maxn]; int vis[maxn]; int main() { #ifdef ONLINE_JUDGE #else freopen("1.txt", "r", stdin); #endif int n,s,id,k; cin>>s>>n>>k; for(int i=0; i<n; i++) { cin>>id; node[id].address=id; cin>>node[id].key>>node[id].next; } vector<int> v[3]; for(int i=s;i!=-1;i=node[i].next){ if(node[i].key<0){ v[0].push_back(i); }else if(node[i].key>k){ v[2].push_back(i); }else v[1].push_back(i); } int flag=0; for (int i = 0; i < 3; i++) { for (int j = 0; j < v[i].size(); j++) { if (flag == 0) { printf("%05d %d ", v[i][j], node[v[i][j]].key); flag = 1; } else { printf("%05d\n%05d %d ", v[i][j], v[i][j], node[v[i][j]].key); } } } printf("-1"); return 0; }