pta——电话聊天狂人(c二叉树实现)
给定大量手机用户通话记录,找出其中通话次数最多的聊天狂人。
输入格式:
输入首先给出正整数N(≤105),为通话记录条数。随后N行,每行给出一条通话记录。简单起见,这里只列出拨出方和接收方的11位数字构成的手机号码,其中以空格分隔。
输出格式:
在一行中给出聊天狂人的手机号码及其通话次数,其间以空格分隔。如果这样的人不唯一,则输出狂人中最小的号码及其通话次数,并且附加给出并列狂人的人数。
输入样例:
4
13005711862 13588625832
13505711862 13088625832
13588625832 18087925832
15005713862 13588625832
输出样例:
13588625832 3
思路:二叉树实现足已,hash可能更快不过用的不熟这里没用。
代码:
#include <stdio.h>
#include <stdlib.h>
typedef long long ll;
typedef struct Node* node;
struct Node {
ll pNumber;
ll number;
node Left;
node Right;
};
node nodeCreate(ll phoneNumber){
node p = (node)malloc(sizeof(struct Node));
p->number = 1;
p->pNumber = phoneNumber;
p->Left = NULL;
p->Right = NULL;
return p;
}
node Search(node root,ll phoneNumber){
if(root == NULL)return root;
node r = root;
while(root){
if(root->pNumber > phoneNumber){
r = root;
root = root->Left;
}
else if(root->pNumber < phoneNumber){
r = root;
root = root->Right;
}
else if(root->pNumber == phoneNumber)return root;
}
return r;
}
node Make(node root,ll phoneNumber){
if(root == NULL){
root = nodeCreate(phoneNumber);
}
else{
node p = Search(root,phoneNumber);
if(p->pNumber == phoneNumber)p->number++;
else {
if(p->pNumber>phoneNumber)p->Left = nodeCreate(phoneNumber);
else p->Right = nodeCreate(phoneNumber);
}
}
return root;
}
ll maxx,sum,mNumber;
void findResult(node root){
if(root == NULL)return ;
if(root->number<maxx);
else if(root->number == maxx){
sum++;
if(root->pNumber<mNumber)mNumber = root->pNumber;
}
else if(root->number>maxx){
sum = 1;
maxx = root->number;
mNumber = root->pNumber;
}
findResult(root->Left);
findResult(root->Right);
}
int main(){
node root = NULL;
int N;
scanf("%d",&N);
while(N--){
ll a,b;
scanf("%lld %lld",&a,&b);
root = Make(root,a);
root = Make(root,b);
}
findResult(root);
printf("%lld %lld",mNumber,maxx);
if(sum>1)printf(" %lld",sum);
return 0;
}