1133. Splitting A Linked List (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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:
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 思路 逻辑水题，将一个链表划分成三个区间。 代码
#include<iostream>
#include<vector>
using namespace std;
class node
{
public:
int val;
int next;
};
vector<node> nodes(100000);
int main()
{
vector<vector<int>> res(3);
int start,N,K;
while(cin >> start >> N >> K)
{
for(int i = 0;i < N;i++)
{
int tmp;
cin >> tmp;
cin >> nodes[tmp].val >> nodes[tmp].next;
}
while(start != -1)
{
if(nodes[start].val < 0)
res[0].push_back(start);
else if(nodes[start].val >= 0 && nodes[start].val <= K)
res[1].push_back(start);
else
res[2].push_back(start);
start = nodes[start].next;
}
int cnt = 0;
for(int i = 0;i < res.size();i++)
{
for(int j = 0;j < res[i].size();j++)
{
if(cnt++ == 0)
{
printf("%05d %d ",res[i][j],nodes[res[i][j]].val);
}
else
{
printf("%05d\n%05d %d ",res[i][j],res[i][j],nodes[res[i][j]].val);
}
}
}
printf("-1");
}
}