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.

思路 (双指针)

遍历两个LinkedList, 找到各自的长度, 然后让长的向前走n个节点使得长短相等, 然后共同向前, 相交的位置就是intersection

复杂度

时间O(m+n) 空间O(a)

代码

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        int lena = getLength(headA);
        int lenb = getLength(headB);
        ListNode nodeA = headA;
        ListNode nodeB = headB;
        if (lena > lenb) {
            for (int i =0; i <lena-lenb; i++) {
                nodeA = nodeA.next;
            }
        } else {
            for (int i =0; i <lenb-lena; i++) {
                nodeB = nodeB.next;
            }
        }
        while (nodeA != null && nodeB!= null) {
            if (nodeA == nodeB) {
                return nodeA;
            }
            nodeA = nodeA.next;
            nodeB = nodeB.next;
        }
        return null;
        
    }

    private int getLength(ListNode head) {
        int res = 0;
        while (head != null) {
            head = head.next;
            res++;
        }
        return res;
    }
}