1133. Splitting A Linked List (25)
2017-09-21 13:31
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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.
Input Specification:
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.
Output Specification:
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.
Sample Input:
Sample Output:
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.
Input Specification:
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.
Output Specification:
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.
Sample 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
Sample Output:
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
#include <iostream> #include <vector> #include <cstdio> #include <algorithm> using namespace std; struct xnode { int add,val,next; }; xnode node[100000]; vector<xnode> p,q,r,ans; int head,n,k; int main() { cin>>head>>n>>k; for(int i=0;i<n;++i) { int add,val,next; cin>>add>>val>>next; node[add].add=add; node[add].val=val; node[add].next=next; } int x=head; while(x!=-1) { if(node[x].val<0) p.push_back(node[x]); else if(node[x].val>=0&&node[x].val<=k) q.push_back(node[x]); else r.push_back(node[x]); x=node[x].next; } ans=p; for(auto ln:q) ans.push_back(ln); for(auto ln:r) ans.push_back(ln); int al=ans.size(); for(int i=0;i<al-1;++i) printf("%05d %d %05d\n",ans[i].add,ans[i].val,ans[i+1].add); printf("%05d %d -1\n",ans[al-1].add,ans[al-1].val); return 0; }
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