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.

/** * 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) {        if( headA==null || headB==null) return null;        int lenA = getLen(headA);        int lenB = getLen(headB);        ListNode pa = headA;        ListNode pb = headB;        int dis = Math.abs(lenA - lenB);        if(lenA > lenB){            for(int i = 0; i < dis; i++){                pa = pa.next;            }        } else if(lenB > lenA){            for(int i = 0; i < dis; i++){                pb = pb.next;            }        }        while(pa != null && pa.val != pb.val){            pa = pa.next;            pb = pb.next;        }        return pa;    }    public int getLen(ListNode head){        int len = 0;        for(ListNode n = head; n != null; n = n.next){            len ++;        }        return len;    }}

思路：

1、用栈。将两条链表存入栈中，然后逐个pop比较。需消耗较多空间，不能满足题目要求。

2、快行指针思想，遍历时，沿较长的链表先走。