1074 Reversing Linked List (25 分)
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.
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 (≤) which is the total number of nodes, and a positive K (≤) which is the length of the sublist to be reversed. 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, and Next is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
Sample Output:
00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1题意:
给N个链表结点，以及K，对每K个长度的链表做逆置，输出逆置后的链表。
题解:
不是很熟悉vector.reverse()这种工具，用了两个双端队列和一个栈来实现。最后一个测试点没过，原来是有些节点不在链表上，那么需要重新计算节点个数，加个计数器sum，不一定就是n。
AC代码:
#include<iostream>
#include<algorithm>
#include<deque>
#include<stack>
using namespace std;
struct node{
int v;
int zhi;
int nx;
}a[100005];
deque<node>q1,q2;
stack<node>st;
int main(){
int root,n,k;
cin>>root>>n>>k;
for(int i=1;i<=n;i++){
int x;
cin>>x;
cin>>a[x].v>>a[x].nx;
a[x].zhi=x;//把它自己的编号也要记录下来
}
int sum=0;//可能有些节点不在链表上,要重新数
int p=root;
q1.push_back(a[p]);
sum++;
while(a[p].nx!=-1){
int next=a[p].nx;
q1.push_back(a[next]);
sum++;
p=next;
}
for(int i=1;i<=sum/k;i++){
int c=0;
while(!q1.empty()){//k个k个分别装入栈里倒一倒再取出来
node x=q1.front();
q1.pop_front();
st.push(x);
c++;
if(c==k) break;
}
while(!st.empty()){//倒着再取出来
q2.push_back(st.top());
st.pop();
}
}
while(!q1.empty()){
node x=q1.front();
q1.pop_front();
q2.push_back(x);
}
while(!q2.empty()){//输出
node x=q2.front();
q2.pop_front();
if(!q2.empty()) printf("%05d %d %05d
",x.zhi,x.v,q2.front().zhi);
else printf("%05d %d -1",x.zhi,x.v);
}
return 0;
}别人的代码：
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;
struct node{
int add,data,Next;
}a[maxn];
vector<node>valid,ans;
int head,n,k;
int main()
{
for(int i=0;i<maxn;i++)a[i].add=i;
scanf("%d%d%d",&head,&n,&k);
for(int i=0;i<n;i++)
{
int address;
scanf("%d",&address);
scanf("%d%d",&a[address].data,&a[address].Next);
}
int p=head;
while(p!=-1)
{
valid.push_back(a[p]);
p=a[p].Next;
}
int group=valid.size()/k;
for(int i=0;i<group;i++)
{
reverse(valid.begin()+i*k,valid.begin()+i*k+k);
}
for(int i=0;i<valid.size();i++)
{
if(i!=valid.size()-1)printf("%05d %d %05d
",valid[i].add,valid[i].data,valid[i+1].add);
else printf("%05d %d -1
",valid[i].add,valid[i].data);
}
return 0;
}