#160 Intersection of Two Linked Lists

题目：

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.

题解：

先跑一下两个链表，得到分别的长度（此时，可以顺便判断下尾节点是否相同，若不同，则不相交，返回null）。

然后长的链表先走m-n（长度差），再两个链表同步走，寻找公共节点。

/** * 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 = 1;        int lenB = 1;        int num = 0;        ListNode tmpA = headA;        ListNode tmpB = headB;        ListNode tmp = null;        while(tmpA.next != null){            tmpA=tmpA.next;            lenA++;        }        while(tmpB.next != null){            tmpB=tmpB.next;            lenB++;        }        if(tmpA != tmpB) return null;                //tmpA为长的链表头结点        if(lenA>=lenB){            tmpA = headA;            tmpB = headB;        }        else{            tmpA = headB;            tmpB = headA;        }        //长链表先走|lenA-lenB|步，以补齐        while(num<Math.abs(lenA-lenB)){            tmpA = tmpA.next;            num++;        }        while(tmpA != tmpB){            tmpA = tmpA.next;            tmpB = tmpB.next;        }        return tmpA;    }}

代码虽长，流程思路还是很简单的。

PS：再吐槽下leetcode的Runtime，同一个代码，第一次运行304ms，第二次524ms，这么任性真的好么！