[C++知识库]c++编程常考知识点

矩阵深层复制（将一个矩阵复制到另一个矩阵里）由于对象不可以直接赋值所以需要重载运算符，

delete p[];
n = r.n;
p=new int [n];
for(int i=0;i<r.n;i++)
{
this->p[i] = r.p[i];
}

两个矩阵相加

Matrix operator+(const Matrix &m1;consr Matrix  &m2)
{
 Matrix NewMatrix;
 int n=0;
 for(int i=0;i<M;i++)
 for(int j=0;j<N;j++)
  {
  n  = m1.getElement(i,j)+m2.getElement(i,j);
  NewMatrix.setElement(i,j,n);
  }
  return NewMatrix;
}

实现矩阵的行列转换

void Matirx::transpose()
{ for(int i=0;i<M;i++)
   for(int j=i+1;j<N;j++)
   {
  int t1=getElement(i,j);
  int t2 =getElement(j,i);
  setElement(i,j,t2);
  setElement(j,i,t1);
   }
}

计数一串字符串中有几个单词

int　count =0;
unsigned i=0;
for(i=0;i<strlen(str);i++)
{
if(str[i]==" "')
count=count+1;
}
if(str[i-1]==" "&& str[i]==''\0')
count=count+1;
return count;

往字符串中插入另一串字符串

MinString & MinString::insert(int pos,char *p,int n)
{ 
   MinString  temp(str);
   delete[] str;
   int OldLen;
  OldLen =strlen（temp.str）;
  str = new char [OldLen+n+1];
  strncpy（str,temp.str,pos）;
  str[pos]='\0';
  //赋值旧字符串到str中
  strncat(str,p,n);
  str[pos+n]='\0';
  //将新字符串插入
  strcat(str,temp.str+pos);
  str[OldLen+n]='\0';
  //将旧字符串pos位置之后的字符插入
    return *this 
}

字符串合并排序

d = new double[len];
while(c < newlen && b < list.len)
{
	if (newdata[c] < list.d[b])
	{
		d[m] = newdata[c];
		m++;
		c++;
	}
	else
	{
		d[m] = list.d[b];
		m++;
		b++;
	}
}

if (c > newlen)
{
	for (; b < list.len; b++)
	{
		d[m] = list.d[b];
		m++;
	}
}
else
{
	for (; c < newlen; c++)
	{
		d[m] = newdata[c];
		m++;
	}
}
delete[]newdata;
}

向有序数列中插入单个数值

void IntegerSet::add(int element)
{
	int j;
	for (j = counter; j > 0; j--)
	{
		if (element >= elem[j - 1])
			break;
	if (j > 0)
		if (element == = elem[j - 1])
			return;
	}
	for (int k = counter; k > j; k--)
		{
		elem[k] = elem[k - 1]；
		elem[j] = element;
	    counter++;
	    }
}

有序数列中每次插入任意个数字

void SortedList::insert(double data)
{
    //判断原数组中是否有插入元素
	for (int i = 0; i < len; i++)
	{
		if (d[i] == data)
			return;
			
		int newlen = len;
		len++;
		double* newdata;
		newdata = new double[newlen];
		//借助中间数组储存原有的数组
		for (intc = 0; c < newlen; c++)
		{
			newdata[c] = d[c];
		}
		
		int j = 0;
		int i = 0;
		delete[]d;
		将原有的空间释放
		d = new double[len];
		while (j < len)
			if (data > newdata[i])
			{
				d[j] = newdata[i];
				j++;
				i++;
			}
			else
			{
				d[j] = data;
				j++;
				for (int a = i; a < len - 1; a++)
				{
					d[j++] = newdata[a];
				}
				return;
			}
	}
}

字符串翻转

int len = strlen(str);
char ch = ' ';
for (int i = 0; i < len / 2; i++)
{
	ch = str[i];
	str[i] = str[len - i - 1];
	str[len - i - 1] = ch;
}
str[len] = '\0';

统计字符串中字母出现的次数

void doc::count()
{
	int i, j;
	for (j = 0; j < length; j++)
	{
		int(i = 0; i < 26; i++)
		{
			if (str[j] == i + 0x41 || str[j] == i + 0x61)
			{
				counter[i]++;
				break;
			}
		}
	}
}

字符串完成重载+

MinString operator+(const MinString& AddStr)
{
	MinString NewStr("");
	if (AddStr.length != 0)
	{
		if (NewStr.sPtr != 0)
		delete[]NewStr.sPtr;
		NewStr.length = length + AddStr.length;
		NewStr.sPtr = new char[NewStr.length + 1];
		strcpy(NewStr.sPtr, AddStr.sPtr);
	}
	return NewStr;
}

数组中数字各个数字之和是偶数

void Sample::calValue(Sample& s)
{
	int qw, bw, sw, gw;
	long he = 0;
	for (int i = 0; i < Max; i++)
	{
		qw = s.A[i] / 1000;
		bw = (s.A[i] % 1000) / 100;
		sw = (s.A[i] % 100) / 10;
		gw = (s.a[i] % 10);
		if ((qw + bw + sw + gw) % 2 == 0)
		{
			s.totCnt++;
			he = he + s.A[i];
		}
	}
	s.totPjz = (double)he / s.totCnt;
}

200个数字求前二十个数字平均数并输出

void Sample::funSort(Sample &s)
{    double value =0.0;
    int i,j,cnt=0;
for(i;i<Max;i++)
{
  cnt++;
  val += s.A[i];
  if(cnt%20==0)
  {
  s.B[j++] = val/20.;
  val = 0;
  }
}    
       for(i=0;i<MAX;i++)
       for(j=i+1;j<Max;j++)
     {
     if(s.B[i]>s.A[j])
     {
         val = s.B[i];
         s.B[i] = s.A[j];
         s.A[j] = val;
              }
     }
}

对数列进行排序，然后将前十个输出

void Sample::funSort(Sample& s)
{
	int i, j, k, temp;
	for (i = 0; i < Max; i++)
	{
		k = i;
		for (j = i + 1; j < Max; j++)
		{
			if (s.A[j] < s.A[k])
				k = j;
			temp = s.A[i];
			s.A[i] = s.A[k];
			s.A[k] = temp;
		}
	}
}

将数组从小到大进行排序

unsigned int i, j, temp;
for (i = 0; i < size - 1; i++)
{
	for (j = i + 1; j < size; j++)
	{
		if (data[i] > data[j])
		{
			temp = data[i];
			data[i] = data[j];
			data[j] = temp;
		}
	}
}