C + + written interview (algorithm problem set III)

1> writing strcpy function, known function prototype char*strcpy (char* strdest,char* strsrc)

ANSWER:

chat* strcpy (char* strdest,char* strsrc)

{

If (Strsrc==null) return NULL;

Char*ch1=strsrc,*ch2=strdest;

while (*ch1!= ')

{

*ch2++=*ch1++;

}

*ch2= ' + ';

Return strdest;

}

2> use recursive method to determine the integer group A[n] is not ascending sort

ANSWER:

Boolean isascending (int a[])

{

Returnisascending (a,0);

}

Bool isascending (int a[],int start)

{

returnstart==length-1 | | Isascending (a,start+1);

}

3> Delete a number in a string and compress a string

CHAR * Partition (const char *STR)

{

char* I=str;

char* J=str;

while (*i!= ')

{

If (*i> ' 9 ' | | | *i< ' 0 ')

*j++=*i++;

Else

*i++;

}

*j= ' + ';

Return str;

}

The 4> function moves the character ' * ' of the string to the front of the string, and the preceding non-' * ' character moves to the back part of the string.

ANSWER:

Int Partitionstr (char a[])

{

Int count=0;

Inti=a.length-1,j=a. Length-1;

while (i>=0)

{

If (a[i]!= ' * ')

{

Swap (a,i--, j--)

}

Else

{

i--;

count++;

}

}

Return count;

}

5> is known as a string that looks for the number of occurrences of the string sub in the original string.

ANSWER:

Int count_of_substr (chat *str,char *sub)

{

Int N=strlen (sub);

Int count=0;

Char *p=stt;

while (p!= ')

{

If (strcmp (p,sub,n))

count++;

p++;

}

Return count;

}

6> A minimal heap, also a fully binary tree, is represented by a layer traversal array.

1, how to access the child nodes of node A[n]

2, insert a node of program void add_element (int *a,int size,int val);

3, remove the minimum node of the program

Anwser:

Void add_element (int *a,int size,int val)

{

A[size]=val;

Int p=size/2-1;

Int c=size;

while (p>=0)

{

If (A[p]<a[c])

break;

Else

{

A[C]=A[P];

C=p;

p= (p-1)/2;

}

}

A[c]=val;

}

Void Del (int *a,int size)

{

A[0]=A[SIZE-1];

Int Val=a[0];

size--;

Int p=0;

Int c=2*p+1;

while (c<=size-1)

{

If (c<size-1 &&a[c]>a[c+1])

C + +;

If (A[p]<=a[c])

break;

Else

{

A[P]=A[C];

P=c; c=2*p+1;

}

}

A[p]=val;

}

7> a combination of strings

Anwser:

Void combination_m (chat *pstr,intm,vector<char> &result)

{

If (pstr==null| | (*pstr== ' &&m!=0))

Return;

If (m==0)

cout<<result<<endl;

Result.push_back (*PSTR);

Combination_m (Pstr+1,m-1,result);

Result.pop_back ();

Combination_m (Pstr+1,m,result);

}

Void combination (char *pstr)

{

If (pstr==null| | *pstr== ' + ')

Return;

Intlen=strlen (str);

for (inti=1;i<=len;i++)

{

vector<char> result;

Combination_m (PSTR,I,RESULR);

}

}