我有http://www.cplusplus.com/forum/beginner/73928/
上定义的完全相同的Elem
和List
类
如果所有值都重复了两次或更多次,您能否建议一些关于如何编写返回true
的函数的提示? E.g。
1,1,1,2,2 - true
1,2 - false
我觉得它肯定需要一个动态数组但却无法想到算法。
答案 0 :(得分:2)
是的,制作一个std::map<int,int>
,用于计算列表中每个数字的出现次数。这个计算需要一次遍历所有列表。
然后,再次对你刚刚创建的std::map
进行传递,找出所有值是否大于或等于2.
答案 1 :(得分:0)
该函数看起来像这样(未经测试):
std::map<int,int> m_mapCount;
std::map<int,int>::iterator m_Iterator;
for (l.start(); !l.end(); l.next()) // put the content of your linkedlist to map
{
m_mapCount[l.current->num] += 1;
}
for (m_Iterator=m_mapCount.begin(); m_Iterator!=m_mapCount.end(); m_Iterator++)
{
if(m_Iterator->second >= 2) return true;
}
答案 2 :(得分:0)
bool twoormore()
{
int count = 0;// for counting elements in list
int temp;// temprorary element for sorting and logical part
int cik;// how much times the value has been mentioned
bool res = true;// function result
int * arr;// pointer for the upcoming dynamic array
for(start();!end();next())
{
count++;// counting the elements
}
if(count != 0){
arr = new int[count];//creating array
int i = 0;
for(start();!end();next())
{
arr[i++] = current->num;//filling array
}
/** array sorting **/
for(int i = 0;i < count;i++)
for(int j = 0; j < count; j++)
{
if(arr[j] > arr[i])
{
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
/** sort ends **/
temp = arr[0]; // setting first element ar temp.. for upcoming check
cik = 1;// it's been its first time
for(int i = 1;i < count;i++)
{
if(arr[i] == temp)
{
cik++; continue;// if upciming element is equal to temprorary , then add 1 to counter.. and continue looping
}else
{
if(cik > 1)
{
temp = arr[i];// if everything ok, but element value changes.
cik = 1;// sets defualt
continue;
}
else
{
res = false;// other way, the value wasnt there two times
break;
}
}
}
delete arr;//deleting allocated space for array
return res;// returning bool, true or false.
}
}