Sunday, November 29, 2015

Sum of Two Linked Lists - 1,2 - 我的博客 - ITeye技术网站


Sum of Two Linked Lists - 1 - 我的博客 - ITeye技术网站
Given two numbers represented by two lists, write a function that returns sum list. The sum list is list representation of addition of two input numbers.
Example 1
Input:
  First List: 5->6->3  // represents number 365
  Second List: 8->4->2 //  represents number 248
Output
  Resultant list: 3->1->6  // represents number 613
Example 2
Input:
  First List: 7->5->9->4->6  // represents number 64957
  Second List: 8->4 //  represents number 48
Output
  Resultant list: 5->0->0->5->6  // represents number 65005
Solution
Traverse both lists. One by one pick nodes of both lists and add the values. If sum is more than 10 then make carry as 1 and reduce sum. If one list has more elements than the other then consider remaining values of this list as 0. Following is Java implementation of this approach.

  1. public ListNode addLinkedList(ListNode a, ListNode b) {  
  2.     ListNode result = new ListNode(0);  
  3.     ListNode dummy = result;  
  4.     int carry = 0;  
  5.     while(a!=null || b!=null) {  
  6.         int sum = carry + ((a!=null)?a.val:0) + ((b!=null)?b.val:0);  
  7.         carry = sum / 10;  
  8.         result.next = new ListNode(sum%10);  
  9.         result = result.next;  
  10.         if(a!=null) a=a.next;  
  11.         if(b!=null) b=b.next;  
  12.     }  
  13.     if(carry>0) {  
  14.         result.next = new ListNode(carry);  
  15.     }  
  16.     return dummy.next;  
http://yuanhsh.iteye.com/blog/2181045
iven two numbers represented by two linked lists, write a function that returns sum list. The sum list is linked list representation of addition of two input numbers. It is not allowed to modify the lists. Also, not allowed to use explicit extra space (Hint: Use Recursion).
Example
Input:
  First List: 5->6->3  // represents number 563
  Second List: 8->4->2 //  represents number 842
Output
  Resultant list: 1->4->0->5  // represents number 1405
Following are the steps.
1) Calculate sizes of given two linked lists.
2) If sizes are same, then calculate sum using recursion. Hold all nodes in recursion call stack till the rightmost node, calculate sum of rightmost nodes and forward carry to left side.
3) If size is not same, then follow below steps:
….a) Calculate difference of sizes of two linked lists. Let the difference be diff
….b) Move diff nodes ahead in the bigger linked list. Now use step 2 to calculate sum of smaller list and right sub-list (of same size) of larger list. Also, store the carry of this sum.
….c) Calculate sum of the carry (calculated in previous step) with the remaining left sub-list of larger list. Nodes of this sum are added at the beginning of sum list obtained previous step.
  1. private int carry = 0;  
  2.   
  3. public ListNode addReversedLinkedList(ListNode a, ListNode b) {  
  4.     if(a==null) {  
  5.         return b;  
  6.     } else if(b==null) {  
  7.         return a;  
  8.     }  
  9.     int m = this.getListSize(a);  
  10.     int n = this.getListSize(b);  
  11.     ListNode result;  
  12.     if(m==n) {  
  13.         result = this.addListWithSameSize(a, b);  
  14.     } else {  
  15.         int diff = Math.abs(m-n);  
  16.         ListNode list1 = m>n?a:b; // make sure the first list is larger  
  17.         ListNode list2 = m>n?b:a;   
  18.         ListNode cur = list1;  
  19.         while(diff-->0) cur = cur.next;  
  20.         result = this.addListWithSameSize(cur, list2);  
  21.         result = this.addRemainingList(list1, cur, result);  
  22.     }  
  23.     if(this.carry>0) {  
  24.         ListNode newRes = new ListNode(0);  
  25.         newRes.val = this.carry;  
  26.         newRes.next = result;  
  27.         return newRes;  
  28.     }  
  29.     return result;  
  30. }  
  31.   
  32. private int getListSize(ListNode n) {  
  33.     int size = 0;  
  34.     while(n!= null) {  
  35.         n = n.next;  
  36.         size++;  
  37.     }  
  38.     return size;  
  39. }  
  40.   
  41. private ListNode addListWithSameSize(ListNode a, ListNode b) {  
  42.     if(a==null || b==nullreturn null;  
  43.     ListNode result = new ListNode(0);  
  44.     result.next = addListWithSameSize(a.next, b.next);  
  45.     int sum = a.val+b.val+this.carry;  
  46.     this.carry = sum/10;  
  47.     result.val = sum%10;  
  48.     return result;  
  49. }  
  50.   
  51. private ListNode addRemainingList(ListNode list1, ListNode cur, ListNode result) {  
  52.     if(list1==cur) return result;  
  53.     ListNode next = addRemainingList(list1.next, cur, result);  
  54.     int sum = list1.val + this.carry;  
  55.     this.carry = sum/10;  
  56.     ListNode ret = new ListNode(0);  
  57.     ret.val = sum%10;  
  58.     ret.next = next;  
  59.     return ret;  
  60. }  
Time Complexity: O(m+n) where m and n are the sizes of given two linked lists.
int getSize(struct node *node)
{
    int size = 0;
    while (node != NULL)
    {
        node = node->next;
        size++;
    }
    return size;
}
 
// Adds two linked lists of same size represented by head1 and head2 and returns
// head of the resultant linked list. Carry is propagated while returning from
// the recursion
node* addSameSize(node* head1, node* head2, int* carry)
{
    // Since the function assumes linked lists are of same size,
    // check any of the two head pointers
    if (head1 == NULL)
        return NULL;
 
    int sum;
 
    // Allocate memory for sum node of current two nodes
    node* result = (node *)malloc(sizeof(node));
 
    // Recursively add remaining nodes and get the carry
    result->next = addSameSize(head1->next, head2->next, carry);
 
    // add digits of current nodes and propagated carry
    sum = head1->data + head2->data + *carry;
    *carry = sum / 10;
    sum = sum % 10;
 
    // Assigne the sum to current node of resultant list
    result->data = sum;
 
    return result;
}
 
// This function is called after the smaller list is added to the bigger
// lists's sublist of same size.  Once the right sublist is added, the carry
// must be added toe left side of larger list to get the final result.
void addCarryToRemaining(node* head1, node* cur, int* carry, node** result)
{
    int sum;
 
    // If diff. number of nodes are not traversed, add carry
    if (head1 != cur)
    {
        addCarryToRemaining(head1->next, cur, carry, result);
 
        sum = head1->data + *carry;
        *carry = sum/10;
        sum %= 10;
 
        // add this node to the front of the result
        push(result, sum);
    }
}
 
// The main function that adds two linked lists represented by head1 and head2.
// The sum of two lists is stored in a list referred by result
void addList(node* head1, node* head2, node** result)
{
    node *cur;
 
    // first list is empty
    if (head1 == NULL)
    {
        *result = head2;
        return;
    }
 
    // second list is empty
    else if (head2 == NULL)
    {
        *result = head1;
        return;
    }
 
    int size1 = getSize(head1);
    int size2 = getSize(head2) ;
 
    int carry = 0;
 
    // Add same size lists
    if (size1 == size2)
        *result = addSameSize(head1, head2, &carry);
 
    else
    {
        int diff = abs(size1 - size2);
 
        // First list should always be larger than second list.
        // If not, swap pointers
        if (size1 < size2)
            swapPointer(&head1, &head2);
 
        // move diff. number of nodes in first list
        for (cur = head1; diff--; cur = cur->next);
 
        // get addition of same size lists
        *result = addSameSize(cur, head2, &carry);
 
        // get addition of remaining first list and carry
        addCarryToRemaining(head1, cur, &carry, result);
    }
 
    // if some carry is still there, add a new node to the fron of
    // the result list. e.g. 999 and 87
    if (carry)
        push(result, carry);
}
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