1097.Deduplication on a Linked List (25)

1097.Deduplication on a Linked List (25)

pat-al-1097

2017-03-03

使用order来记录是第几个结点，一个巧妙的方式分开了需要保留下来的结点和需要摘除的结点

/**
* pat-al-1097
* 2017-03-03
* Cpp version
* Author: fengLian_s
*/
#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
#define MAX 100010
using namespace std;
struct NODE
{
int addr;
int key;
int next;
int order;
}node[MAX];
bool cmp(NODE a, NODE b)
{
return a.order < b.order;
}
int isExist[MAX] = {0};
int main()
{
freopen("in.txt", "r", stdin);
int start, n;
scanf("%d%d", &start, &n);
for(int i = 0;i < MAX;i++)
{
node[i].order = 2*MAX;
}
for(int i = 0;i < n;i++)
{
int address;
scanf("%d", &address);
scanf("%d %d", &node[address].key, &node[address].next);
node[address].addr = address;
//printf("%05d %d %05d %d\n", node[address].addr, node[address].key, node[address].next, node[address].order);
}
int index = start, orderResult = 0, orderRemove = 0;
while(index != -1)
{
if(isExist[abs(node[index].key)] == 0)
{
isExist[abs(node[index].key)] = 1;
node[index].order = orderResult++;
}
else
node[index].order = MAX + orderRemove++;
//printf("%05d %d %05d %d\n", node[index].addr, node[index].key, node[index].next, node[index].order);
index = node[index].next;
}
sort(node, node+MAX, cmp);
int cnt = orderResult+orderRemove;
for(int i = 0;i < cnt;i++)
{
//printf("%05d %d %05d %d\n", node[i].addr, node[i].key, node[i].next, node[i].order);
if(i != orderResult-1 && i != cnt-1)
printf("%05d %d %05d\n", node[i].addr, node[i].key, node[i+1].addr);
else
printf("%05d %d -1\n", node[i].addr, node[i].key);
}
return 0;
}

-FIN-