求一个数组中出现次数最多的数

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描叙：一大推数据里面，数字与数字之间用空格隔开，找出出现次数最多的一个数字的算法

#include<stdio.h>void FindMostTimesDigit(int *Src , int SrcLen){int element , has = SrcLen;int MaxNum , TempCount = 0 , MaxCount = 0;int i , j;int *result = new int[];while(has != 0){element = Src[has - 1];TempCount = 0;for(i = has - 1 ; i >= 0 ; --i){//如果找到，则计数加1 ，然后将数据和末尾交换//这也是为何要从末尾开始循环的理由if(element == Src[i]){TempCount++;Src[i] = Src[has - 1];has--;}}if(TempCount > MaxCount){MaxCount = TempCount;MaxNum = 0;result[MaxNum] = element;}else if(TempCount == MaxCount){result[++MaxNum] = element;}}printf("出现次数最多的次数：%d\n" , MaxCount);for(j = 0 ; j <= MaxNum ; ++j)printf("%d " , result[j]);printf("\n");}int main(){int list[] = {1,2,3,4,3,3,2,2,1,1,4,4,4,1,2};int length = sizeof(list) / sizeof(int);FindMostTimesDigit(list , length);return 0;}

C++解法如下：

#include<iostream>#include<map>#include<utility>using namespace std;int main(){map<int , int> word_count;int number;while(cin>>number){pair<map<int , int>::iterator , bool> ret = word_count.insert(make_pair(number , 1));if(!ret.second)++ret.first->second;}for(map<int , int>::iterator iter = word_count.begin() ; iter != word_count.end() ; ++iter)cout<<(*iter).first<<"\t\t"    <<(*iter).second<<endl;return 0;}

更简洁的方法如下：

#include<iostream>#include<map>#include<utility>using namespace std;int main(){map<int , int> word_count;int number;while(cin>>number)++word_count[number];for(map<int , int>::iterator iter = word_count.begin() ; iter != word_count.end() ; ++iter)cout<<(*iter).first<<"\t\t"    <<(*iter).second<<endl;return 0;}