Leetcode 160. Intersection of Two Linked Lists

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.

Analysis:

3 steps:

• Get the length of both list
• Let the longer list to go before len1-len2 steps
• Make the lists go further, if they’re the same, then we find it

Time complexity: O(m + n)

Space complexity: O(1)

Code is below:

/**
 * 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) {
        // O(m + n)
        
        if (headA == null || headB == null) {
            return null;
        }
        
        int lenA = getLength(headA);
        int lenB = getLength(headB);
        
        // Let the longer list to go before lenA - lenB 
        for (int i = 0; i < Math.abs(lenA - lenB); i++) {             if (lenA > lenB) {
                headA = headA.next;
            } else {
                headB = headB.next;
            }
        }
        
        while (headA != null && headB != null) {
            if (headA == headB) {
                return headA;
            }
            headA = headA.next;
            headB = headB.next;
        }
        
        return null;
    }
    
    public int getLength(ListNode head) {
        int len = 0;
        
        while (head != null) {
            head = head.next;
            len++;
        }
        
        return len;
    }
}