1133 Splitting A Linked List (25分)
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 (≤10
5
) which is the total number of nodes, and a positive K (≤10
3
). 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 [−10
5
,10
5
], 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
就是给了一个链表，按照负数调在前面，0-K 在中的 顺序调整，我第一想法是构建一个lsit< node > 然后再这个list表先提取负数，再提取0-k之间的数，剩下的顺序不变
几个关键的点
-
如何保存输入给的结点信息，我是用了两个map 映射，根据pos 映射对应的data和next，也可以直接使用一个node数组，都行
struct node { int data, next; }list[100000]; -
如何将保存的结点信息构造出来，我是用的list< node > ,看了别人的代码感觉更好的办法是建立三个vector< node > 根据data小于0 大于0小于K 大于K 分别保存，这样提取时候也简单，如果使用list 的提取就需要进行三次，每次提取之后使用erase（it++） 删除对应的结点
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <unordered_map>
#include <list>
using namespace std;
struct node
{
int posl;
int data;
int next;
};
int main()
{
int pos,N,K;
cin>>pos>>N>>K;
vector<node> nodeSave;
unordered_map<int,int> map_data;
unordered_map<int,int> map_next;
for(int i=0; i<N; i++)
{
node a;
cin>>a.posl>>a.data>>a.next;
map_data[a.posl]=a.data;
map_next[a.posl]=a.next;
}
list<node> listt;
while(pos!=-1)
{
node a;
a.posl=pos;
a.data=map_data[pos];
a.next=map_next[pos];
pos=a.next;
listt.push_back(a);
}
vector<node> out;
for(auto it=listt.begin(); it!=listt.end();)
{
// cout<<(*it).posl<<' '<<(*it).data<<endl;
if((*it).data<0)
{
out.push_back(*it);
listt.erase(it++);// 后面的++很关键
}
else
it++;
}
for(auto it=listt.begin(); it!=listt.end();)
{
// cout<<(*it).posl<<' '<<(*it).data<<endl;
if((*it).data<=K)
{
out.push_back(*it);
listt.erase(it++);// 后面的++很关键
}
else
it++;
}
for(auto it=listt.begin(); it!=listt.end();it++)
{
out.push_back(*it);
}
//cout<<out[0].posl;
for(int i=0;i<out.size();i++){
if(i==0){
printf("%05d %d ",out[i].posl,out[i].data);
}
else{
printf("%05d\n%05d %d ",out[i].posl,out[i].posl,out[i].data);
}
}
cout<<-1;
return 0;
}
更好办法柳神：https://blog.csdn.net/liuchuo/article/details/78037305