160. Intersection of Two Linked Lists

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.

My Solutions:

方法1：

• Add list A to end of list B and add list B to end of list A

• If they have intersection, it is located at the end of combined list at same index

  • A: a1 -> a2 -> c1 -> c2 -> c3 -> b1 -> b2 -> b3 -> c1 -> c2 -> c3

  • B: b1 -> b2 -> b3 -> c1 -> c2 -> c3 -> a1 -> a2 -> c1 -> c2 -> c3

• Time: O(m + n) [m, n are lenA and lenB]

• Space: O(1)

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {

        if (headA == null || headB == null) return null;
        ListNode a = headA, b = headB;

        while (a != b) {
            a = (a == null ? headB : a.next);
            b = (b == null ? headA : b.next);
        }
        return b;
    }
}

方法2：

• 求出A和B分别长度 (另写一个求长度method)

• 如果lenA > lenB, 先跳过A的一部分直到lenA == lenB相等；反之同理

• 用while循环headA 和headB直到两者相等

Time: O(n); Space: O(1)

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        
        // get lengths first
        int lenA = getListLength(headA);
        int lenB = getListLength(headB);
        
        // find diff
        boolean aIsLonger = lenA - lenB > 0 ? true : false;
        int diff = aIsLonger? lenA - lenB : lenB - lenA;
        
        // move the head of the longer list to the place where lists have same length
        ListNode currA = headA, currB = headB;
        if (aIsLonger) {
            while (diff != 0) {
                currA = currA.next;
                diff--;
            }
        } else {
            while (diff != 0) {
                currB = currB.next;
                diff--;
            }
        }
        
        // now both lists have the same length left
        while (currA != null) {
            if (currA == currB) return currA;
            else {
                currA = currA.next;
                currB = currB.next;
            }
        }
        
        // return null if cannot find a common node
        return null;
    }
    
    private int getListLength(ListNode head) {
        int len = 0;
        while (head != null) {
            len++;
            head = head.next;
        }
        return len;
    }