PAT-Deduplication on a Linked List
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- 题意:给你n条记录,每天记录由一下三部分组成(地址,key,next_address),再给你一个起始地址。让你输出两个新的list,一个是将原来的list删除key的绝对值之后的list,一个是被删除元素组成的list,第一个被删除的元素就是该list 的首元素。
- 解法:我们定义两个数组,一个用来记录key,一个用来记录next address, 数组的index就是当前的address。因为题中说明了address最多10^5,所以开个数组的内存还是可以的。在声明一个数组用来记录key是否出现过,题中也给出了key的范围10^4。最后定义两个vector,一个用来存放第一个list的address,一个用来存放第二个list的address,也就是被删除元素组成的list。
- 代码如下所示:
#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <cmath>using namespace std;const int max_number_address = 99999;int record_key[max_number_address];int record_next[max_number_address];const int max_key = 1e+5;bool exists[max_key];int fabs(int x){ if(x<0) { return -x; } return x;}int main(){ freopen("/home/give/PAT/DeduplicationonaLinkedList.txt","r",stdin); memset(exists, false, sizeof(exists)); int first_address,n; cin>>first_address>>n; for(int i=0;i<n;i++) { int address, key, next; cin>>address>>key>>next; record_key[address] = key; record_next[address] = next; } int cur_address = first_address; vector<int> used; vector<int> unused; for(int i=0;i<n;i++) { int cur_key = record_key[cur_address]; if(exists[fabs(cur_key)] == false) { used.push_back(cur_address); exists[fabs(cur_key)] = true; }else{ unused.push_back(cur_address); } cur_address = record_next[cur_address]; } for(int i=0;i<used.size();i++) { cur_address = used[i]; int cur_value = record_key[cur_address]; int cur_next; if(i==(used.size()-1)) { cur_next = -1; printf("%05d %d %d\n", cur_address, cur_value, cur_next); }else{ cur_next = used[i+1]; printf("%05d %d %05d\n", cur_address, cur_value, cur_next); } } for(int i=0;i<unused.size();i++) { cur_address = unused[i]; int cur_value = record_key[cur_address]; int cur_next; if(i==(unused.size()-1)) { cur_next = -1; printf("%05d %d %d\n", cur_address, cur_value, cur_next); }else{ cur_next = unused[i+1]; printf("%05d %d %05d\n", cur_address, cur_value, cur_next); } }}
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