LeetCode 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.

思路分析：这题考察链表操作，首先要正确理解题意。如果两个链表有intersection，那么最后一个node必然是同一个node，否则就是没有intersection。然后如果我们已经知道两个链表有intersection，如果去找第一个交叉的node呢？最直接的方法是用两个指针同时在两个链表走，但是如何保证它们可以同时到达那个首个交叉的node呢？我们可以让这两个指针的起点不同，保证他们同时走的时候，到达首个交叉的node的距离相等。于是我们想到利用两个链表的长度之差|len1 - len2|.如果len1 > len2,那么就先让pa先走|len1 - len2|步，然后pa和pb一起走，必然同时到达首个交叉的node。对于len1 <= len2的情况，同理可以分析。

AC Code

/** * 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;        ListNode pa = headA;        ListNode pb = headB;        int lenA = 1;        int lenB = 1;        while(pa.next != null){            pa = pa.next;            lenA++;        }        while(pb.next != null){            pb = pb.next;            lenB++;        }        if(pa != pb) return null; //no intersection        pa = headA;        pb = headB;        int diff = Math.abs(lenA - lenB);        if(lenA > lenB){            while(diff != 0 ){                pa = pa.next;                diff--;            }        } else {            while(diff != 0){                pb = pb.next;                diff--;            }        }        while(pa != pb){            pa = pa.next;            pb = pb.next;        }        return pa;    }}