Given two numbers represented as strings, return multiplication of the numbers as a string.
Note: The numbers can be arbitrarily large and are non-negative.

string stringMultip(string &A, string &B) {

	string res(A.size() + B.size(), '0');
	
	string::reverse_iterator rit1, rit2, rit3, rit4;
	int num_a, num_b, num_c, num_carry, num_res, i;
	
	for(rit1 = A.rbegin(), rit3 = res.rbegin(); rit1 != A.rend(); rit1 ++, rit3 ++) {
		
		num_a = (*rit1) - '0';
		num_carry = 0;
		
		for(rit2 = B.rbegin(), rit4 = rit3; rit2 != B.rend(); rit2 ++, rit4 ++) {
			
			num_b = (*rit2) - '0';
			num_c = (*rit4) - '0';
			
			num_res = num_a * num_b + num_c + num_carry;
			num_carry = num_res / 10;
			(*rit4) = '0' + (num_res % 10);
		}

		// put the carry back to the result
		(*rit4) = '0' + num_carry;
	}

	// remove leading '0's
	for(i = 0; i < res.size(); i ++)
		if(res[i] != '0') 
			break;

	return res.erase(0, i);
}

The following provides a simple test case.

#include <iostream>
#include <string>

using namespace std;

int main ()
{
	string A("123456");
	string B("294987");
	string C("123");
	string D("234");	

	cout << stringMultip(A, B) <<endl;
	cout << stringMultip(C, D) <<endl;
	
	return 0;
}

The problem:
Given a binary tree, return the postorder traversal of its nodes’ values. For example, given binary tree:
a              1
b           /      \
c        4           2
d          \       /
e           10  3
return [10, 4, 3, 2, 1]

#include <iostream>
#include <stack>
#include <string>

using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x): val(x), left(nullptr), right(nullptr) {};
};

int postorderTraversal(TreeNode *root) {

    const TreeNode *p = root;
    stack<const TreeNode*> s;

    if(p!=nullptr) {
        TreeNode x(p->val);
        s.push(&x);

        if(p->right != nullptr) s.push(p->right);
        if(p->left != nullptr) s.push(p->left);
    }

    while(!s.empty()) {

        p = s.top();
        s.pop();

        if(p->left == nullptr && p->right == nullptr)
            printf("%d\n", p->val);
        else {
            TreeNode x(p->val);
            s.push(&x);

            if(p->right != nullptr) s.push(p->right);
            if(p->left != nullptr) s.push(p->left);
        }
    }

    return 0;
}

int main() {

    TreeNode root(1);
    TreeNode node1(2);
    TreeNode node2(3);
    TreeNode node3(4);
    TreeNode node4(10);

    root.left = &node3;
    root.right = &node1;
    node1.left = &node2;
    node3.right = &node4;

    postorderTraversal(&root);

    return 0;
}

The similar code structure could be used for binary tree inorder traversal. The key idea here is to create a temporary node which holds only the value and push it into the stack. Whenever we meet a node without both childern, we print out its value.

The Problem:

Given a string containing just the characters ‘(‘ and ‘)’, find the length of the longest valid (well formed) parentheses sub-strings.

#include <iostream>
#include <string>
#include <stack>

using namespace std;

int longest_valid_parentheses(string const& s) {

    int count = 0, max_count = 0;
    stack<char> stk;

    for(auto c : s) {

        if(c == '(')
            stk.push(c);
        else {
            if( stk.empty() ) //|| stk.top() != '(')
                count = 0;
            else {
                count += 2;
                stk.pop();
            }
        }

        max_count = count > max_count ? count : max_count;
    }

    return max_count;
}

You can test the above code using the following test cases @ http://www.cpp.sh/.

int main()
{
  string test1("(()");
  string test2("(()()(");
  string test3(")()()))(((())))");

  cout << "there are " << longest_valid_parentheses(test1) << " valid parentheses in " << test1 << endl;
  cout << "there are " << longest_valid_parentheses(test2) << " valid parentheses in " << test2 << endl;
  cout << "there are " << longest_valid_parentheses(test3) << " valid parentheses in " << test3 << endl;

  return 0;
}

The output would look like as the follows:

there are 2 valid parentheses in (() 
there are 4 valid parentheses in (()()( 
there are 8 valid parentheses in )()()))(((())))

Given a sorted linked list, remove all the duplicates from the list. For example, given 1->2->3->3->6, return 1->2->6. Another example, given 1->1->1->2->4, return 2->4.

typedef struct node Node;

/* The function removes duplicates from a sorted list */
void removeDuplicates(Node** head)
{ 
  /* do nothing if the list is empty */
  Node* current = *head;  
  if(current == NULL)
     return;

  /* Pointer to traverse the linked list */
  Node* pre_head = (Node*) malloc(sizeof(Node));
  pre_head->next = *head;

  /* Pointer to store the next pointer of a node to be deleted*/
  Node* next_next;
  Node* pre = pre_head;
  bool deleted = false;
 
  /* Traverse the list till last node */
  while(current->next != NULL)
  {
   /* Compare current node with next node */
    if(current->data == current->next->data) {       
		/*The sequence of steps is important*/              
		next_next = current->next->next;
		free(current->next);
		current->next = next_next;
		deleted = true;
    }
    else if (deleted)  {
		pre->next = current->next;
		free(current);
		current = pre->next;
		deleted = false;
    }
	else {
		pre = current;
		current = current->next;
    }
  }

  /* remove the last duplicated node*/
  if (deleted) {
	  pre->next = NULL;
	  free(current);
  }
  
  *head = pre_head->next;
  free(pre_head);
}
 

To test the code, please use the testing program from http://www.geeksforgeeks.org/remove-duplicates-from-a-sorted-linked-list/.

Dutch national flag problem is a classic programming problem. It groups the input array into three sub-groups in just one pass. For detailed information, please refer to http://en.wikipedia.org/wiki/Dutch_national_flag_problem

function x = category_sort(x, low, high)

len = length(x);
p = 1; q = len;
i = 1;

while i <= q
    if x(i) < low     % move the current element to the front
        tmp = x(p);
        x(p) = x(i);
        x(i) = tmp;

        p = p + 1;  
        i = i + 1;     
    elseif x(i) >= high     % move the current element to the back
        tmp = x(q);
        x(q) = x(i);
        x(i) = tmp;
                
        q = q - 1;
    else    % move to the next element in the array
        i = i + 1;
    end    
end

We can test the code by simply running the following script:

x = 3 * rand(20, 1);
x = category_sort(x, 1, 2)

To implement the C function ‘strtok’, one doesn’t need to allocate extra memory for the input string before modifying it, since the caller function is supposed to make copy of the input string and deallocate the original input string. The function only needs to define static variables to store the starting of the string.

The caller function provides a string as the input during the first call, and ‘null’ for the following function calls. For details about how this function works, please refer to http://www.cplusplus.com/reference/cstring/strtok/.

char* sp = NULL; /* the start position of the string */

char* strtok1(char* str, const char* delimiters) {

	int i = 0;
	int len = strlen(delimiters);

	/* check in the delimiters */
	if(len == 0)
		printf("delimiters are empty\n");

	/* if the original string has nothing left */
	if(!str && !sp)
		return NULL;

	/* initialize the sp during the first call */
	if(str && !sp)
		sp = str;

	/* find the start of the substring, skip delimiters */
	char* p_start = sp;
	while(true) {
		for(i = 0; i < len; i ++) {
			if(*p_start == delimiters[i]) {
				p_start ++;
				break;
			}
		}

		if(i == len) {
		       sp = p_start;
		       break;
		}
	}

	/* return NULL if nothing left */
	if(*sp == '\0') {
		sp = NULL;
		return sp;
	}

	/* find the end of the substring, and
        replace the delimiter with null */
	while(*sp != '\0') {
		for(i = 0; i < len; i ++) {
			if(*sp == delimiters[i]) {
				*sp = '\0';
				break;
			}
		}

		sp ++;
		if (i < len)
			break;
	}

	return p_start;
}

This following is the test code from cplusplus.com.

#include <stdio.h>
#include <string.h>

int main ()
{
  char str[] ="- This, a sample string.";
  char * pch;
  printf ("Splitting string \"%s\" into tokens:\n",str);
  pch = strtok1(str," ,.-");
  while (pch != NULL)
  {
    printf ("%s\n",pch);
    pch = strtok1(NULL, " ,.-");
  }
  return 0;
}

————————-
As pointed out by Ian in the following comments [9.25.2014], there is a much simpler solution to this problem. Feel free to test it and let us know if it has any bugs.

char* strtok1(char *str, const char* delim) {
	static char* _buffer;
	if(str != NULL) _buffer = str;
	if(_buffer[0] == '\0') return NULL;

	char *ret = _buffer, *b;
	const char *d;

	for(b = _buffer; *b !='\0'; b++) {
		for(d = delim; *d != '\0'; d++) {
			if(*b == *d) {
				*b = '\0';
				_buffer = b+1;

				// skip the beginning delimiters
				if(b == ret) { 
					ret++; 
					continue; 
				}
				return ret;
			}
		}
	}

	return ret;
}

I came to this question because of the Kaiser window used for my BM3D implementation. I used Matlab to generate the Kaiser window I want for my C code so I don’t bother to write that function in my project. However, Matlab only has 1D version of the window functions. The following demonstrates two ways to generate the 2D window from its 1D function.

——————————————————————-

There are basically two ways of forming a 2D window from a 1D function. The
first is the outer product formulation; the second is the rotational
formulation. To create a outer product window (which your kaiser examples
look like they are trying to do), you can use

w1D = hamming(16); % Some 1D window
w2D = w1D(:) * w1D(:).’; % Outer product

To use a rotational formulation (which generally gives more circular
contours), you can use

L = 16;
w1D = hamming(L); % Some 1D window
M = (L-1)/2;
xx = linspace(-M,M,L);
[x,y] = meshgrid(xx);
r = sqrt( x.^2 + y.^2 );
w2D = zeros(L);
w2D(r<=M) = interp1(xx,w1D,r(r<=M));

http://www.mathworks.com/matlabcentral/newsreader/view_thread/23588