描叙：一大堆数据里面，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 , *result = new int[];
while(0 != has)
{
TempCount = 0;
element = Src[has - 1];
for(j = has - 1 ; j >= 0 ; --j)
{
// 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[j])
{
TempCount++;
// 把后面的数据移动到前面来
Src[j] = Src[has - 1];
has--;
}
}
if(TempCount > MaxCount)
{
MaxCount = TempCount;
MaxNum = 0;
result[MaxNum] = element;
}
else if(TempCount == MaxCount)
{
result[++MaxNum] = element;
}
}
printf("出现最多的次数：%d
" , MaxCount);
for(i = 0 ; i <= MaxNum ; ++i)
{
printf("%d " , result[i]);
}
printf("
");
}
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：
1 #include<iostream> 2 #include<map> 3 #include<utility> 4 using namespace std; 5 6 int main() 7 { 8 map<int , int> word_count; 9 int number; 10 while(cin>>number) 11 { 12 pair<map<int , int>::iterator , bool> ret = word_count.insert(make_pair(number , 1)); 13 if(!ret.second) 14 ++ret.first->second; 15 } 16 17 for(map<int , int>::iterator iter = word_count.begin() ; iter != word_count.end() ; ++iter) 18 cout<<(*iter).first<<" " 19 <<(*iter).second<<endl; 20 21 return 0; 22 }
更简洁的方法如下：
1 #include<iostream> 2 #include<map> 3 #include<utility> 4 using namespace std; 5 6 int main() 7 { 8 map<int , int> word_count; 9 int number; 10 11 while(cin>>number) 12 ++word_count[number]; 13 14 for(map<int , int>::iterator iter = word_count.begin() ; iter != word_count.end() ; ++iter) 15 cout<<(*iter).first<<" " 16 <<(*iter).second<<endl; 17 18 return 0; 19 }
网上看到一哥们也写了类似的东西：http://blog.csdn.net/tianmohust/article/details/7514618