本題思路:
遍歷整個鏈表,篩選出兩個結構體數組,一個放本體,另一個放副本.通過bool類型數組中的記錄判斷當前遍歷到的key是本體還是副本,再將其放入對應的結構體中.
注意點:可能存在無效結點.
#include<cstdio>
#include<cstdlib>
#include<string.h>
#include<math.h>
#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=100010;
bool iscur[10010]={0};
struct node{
int address,key,next;
}nodes[maxn],node1[maxn],node2[maxn];
int main(){
#ifdef ONLINE_JUDGE
#else
freopen("1.txt","r",stdin);
#endif
int first,n,idx1=0,idx2=0;
scanf("%d%d",&first,&n);
for(int i=0;i<n;i++){ //初始化輸入
int address;
scanf("%d",&address);
scanf("%d%d",&nodes[address].key,&nodes[address].next);
nodes[address].address=address;
}
while(first!=-1){ //遍歷整個list,分別篩選到兩個結構體數組
int akey=nodes[first].key;
if(akey<0)akey=-akey;
if(iscur[akey]){
node2[idx2++]=nodes[first];
}else{
node1[idx1++]=nodes[first];
iscur[akey]=true;
}
first=nodes[first].next;
}
for(int i=0;i<idx1;i++){ //輸出resulting linked list
printf("%05d %d ",node1[i].address,node1[i].key);
if(i==idx1-1){
printf("-1\n");
}else{
printf("%05d\n",node1[i+1].address);
}
}
for(int i=0;i<idx2;i++){ //輸出removed list
printf("%05d %d ",node2[i].address,node2[i].key);
if(i==idx2-1){
printf("-1\n");
}else{
printf("%05d\n",node2[i+1].address);
}
}
return 0;
}
算法筆記用的方法實現空間佔用少,在一個結構體中實現.
本體結點、副本結點、無效結點三者各自在自己的大範圍內,方便通過排序將它們區分開來.