160. Intersection of Two Linked Lists

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
      // find the length of list 1 and list 2 
      // get the diff
      // start from the same length position 
      // return the first node that's in both lists 
      
      int lenA = getLen(headA);
      int lenB = getLen(headB);
      
      if(lenA > lenB){
        return helper(headB, headA, lenA - lenB);
      }else{
        return helper(headA, headB, lenB - lenA);
      }   
      
    }
  
    private ListNode helper(ListNode short, ListNode long, int diff){
      while(diff > 0){
        long = long.next;
        diff--;
      }
      
      while( short != long ){
        short = short.next;
        long = long.next;
      }
      return short;
      
    }
  
    private int getLen(ListNode head){
      int count = 1;
      while( head != null && head.next != null){
        head = head.next;
        count++;
      }
      return count;
    }
}

 

Intersection of Two Linked Lists


// not mine 
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    int lenA = length(headA), lenB = length(headB);
    // move headA and headB to the same start point
    while (lenA > lenB) {
        headA = headA.next;
        lenA--;
    }
    while (lenA < lenB) {
        headB = headB.next;
        lenB--;
    }
    // find the intersection until end
    while (headA != headB) {
        headA = headA.next;
        headB = headB.next;
    }
    return headA;
}

private int length(ListNode node) {
    int length = 0;
    while (node != null) {
        node = node.next;
        length++;
    }
    return length;
}

 

Write a program to find the node at which the intersection of two singly linked lists begins.

 

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

 

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.