Leetcode 之 Intersection of Two Linked Lists

问题描述：
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.

一开始 的思路是：把俩list都reverse一下，然后挨个遍历，找到第一个值不相同的，把上一个node返回。但是这样改变了原list，没有通过。

正确的思路是先得到俩list的长度，然后先让长的list遍历，直到所剩余的长度和短list长度相同，再开始俩list同时遍历，返回第一个值相同的node

代码如下：

/** * 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) {        ListNode temp=null,pA=headA,pB=headB;        int len1=0,len2=0;        while(pA!=null||pB!=null){            if(pA!=null){                pA=pA.next;                len1++;            }            else{            pB=pB.next;            len2++;}        }        ListNode longNode,shortNode;        if(len1>len2){            longNode=headA;            shortNode=headB;        }        else{            longNode=headB;            shortNode=headA;        }        int diff=Math.abs(len1-len2),i=0;        while(shortNode!=null&&longNode!=null){            if(i>=diff){                if(shortNode.val==longNode.val)return shortNode;                shortNode=shortNode.next;            }            longNode=longNode.next;            i++;        }        return temp;    }}