中国大学MOOC-陈越、何钦铭-数据结构-2018秋——电话聊天狂人

版权声明:我的GitHub:https://github.com/617076674。真诚求星! https://blog.csdn.net/qq_41231926/article/details/84674733

我的中国大学MOOC-陈越、何钦铭-数据结构-2018秋代码仓:https://github.com/617076674/MOOC-DataStructure-2018-Autumn

题目描述:

知识点:map集合的应用

思路:用map集合存储每个人的号码及其通话次数

时间复杂度和空间复杂度均是O(N)。

C++代码:

#include<iostream>
#include<map>
#include<vector>

using namespace std;

int N;
long long num1, num2;
map<long long, int> countMap;
vector<long long> maxCallTimesPeople;

int main(){
	scanf("%d", &N);
	for(int i = 0; i < N; i++){
		scanf("%lld %lld", &num1, &num2);
		countMap[num1]++;
		countMap[num2]++;
	}
	int maxCallTimes = 0;
	for(map<long long, int>::iterator it = countMap.begin(); it != countMap.end(); it++){
		if(it->second > maxCallTimes){
			maxCallTimes = it->second;
		}
	}
	for(map<long long, int>::iterator it = countMap.begin(); it != countMap.end(); it++){
		if(it->second == maxCallTimes){
			maxCallTimesPeople.push_back(it->first);
		}
	}
	if(maxCallTimesPeople.size() == 1){
		printf("%lld %d\n", maxCallTimesPeople[0], maxCallTimes);
	}else{
		printf("%lld %d %d\n", maxCallTimesPeople[0], maxCallTimes, maxCallTimesPeople.size());
	}
	return 0;
} 

C++解题报告:

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转载自blog.csdn.net/qq_41231926/article/details/84674733