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  1. Prepare
  2. Data Structures
  3. Linked Lists
  4. Find Merge Point of Two Lists
  5. Discussions

Find Merge Point of Two Lists

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1214 Discussions

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  • venansiusmt
     + 0 comments

    The key idea to check the the same object is exist in the set

    static int findMergeNode(SinglyLinkedListNode head1, SinglyLinkedListNode head2) {
        Set<SinglyLinkedListNode> nodeSet = new HashSet<>();
        
        SinglyLinkedListNode currh1 = head1;
        SinglyLinkedListNode currh2 = head2;
        while(currh1 != null){
            nodeSet.add(currh1);
            currh1 = currh1.next;
        }
        
        while(currh2 != null){
            if(nodeSet.contains(currh2)){
                return currh2.data;
            }
            currh2 = currh2.next;
        }
        
        return -1;   
    }

  • kamyap2000
     + 0 comments

    OPTIMIZED APPROACH

     static int length(SinglyLinkedListNode temp)
     {
        int len =0;
        while(temp != null)
        {
            len++;
            temp = temp.next;
        }
        return len;
     }
    static int findMergeNode(SinglyLinkedListNode head1, SinglyLinkedListNode head2) {
        
        int len1 = length(head1);
        int len2 = length(head2);
        int mov = 0;
        SinglyLinkedListNode temp = null;
        boolean head1_moved = false;
        boolean head2_moved = false;
        if(len1 > len2)
        {
            mov = len1 - len2;
            temp = head1;
            while(mov-- > 0)
            {
                temp = temp.next;
            }
            head1_moved = true;
        }
        else if(len1 <= len2)
        {
            mov = len2 - len1;
            temp = head2;
            while(mov-- > 0)
            {
                temp = temp.next;
            }
            head2_moved = true;
        }
        
        if(head1_moved)
        {
            SinglyLinkedListNode curr = head2;
            while(temp != null)
            {
                if(curr == temp)
                {
                    return curr.data;
                }
                temp = temp.next;
                curr = curr.next;
            }
        }
        else if(head2_moved){
            SinglyLinkedListNode curr = head1;
            while(temp != null)
            {
                if(curr == temp)
                {
                    return curr.data;
                }
                temp = temp.next;
                curr = curr.next;
            }
        }
        
        
        return -1;

    }

  • kamyap2000
     + 0 comments

    BRUTE FORCE Approach

     static int findMergeNode(SinglyLinkedListNode head1, SinglyLinkedListNode head2) {

        HashMap<SinglyLinkedListNode, Integer> map = new HashMap<>();
        SinglyLinkedListNode temp1 = head1;
        SinglyLinkedListNode temp2 = head2;
        
        while(temp1 != null)
        {
                map.put(temp1, 1);
                temp1 = temp1.next;
            
        }
        while(temp2 != null)
        {
            if(map.containsKey(temp2))
            {
                return temp2.data;
            }
            else{
                map.put(temp2, 1);
                temp2 = temp2.next;
            }
        }
        return -1;
				
    }

  • sarasithjoy
     + 0 comments

    can you help me to integrate this script into my business setup website , i want to use this on my wordpress website

  • ouadochhassan201
     + 0 comments

    solution with JS

    function findMergeNode(headA, headB) {
   let current1 = headA;
    let current2 = headB;

    let length1 = 0;
    let length2 = 0;

    while (current1) {
        length1++;
        current1 = current1.next;
    }

    while (current2) {
        length2++;
        current2 = current2.next;
    }

    current1 = headA;
    current2 = headB;

    if (length1 > length2) {
        for (let i = 0; i < length1 - length2; i++) {
            current1 = current1.next;
        }
    } else if (length2 > length1) {
        for (let i = 0; i < length2 - length1; i++) {
            current2 = current2.next;
        }
    }

    while (current1 && current2) {
        if (current1 === current2) {
            return current1.data; 
        }
        current1 = current1.next;
        current2 = current2.next;
    }

    return null; 
}