leetcode 160. Intersection of Two Linked Lists

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本文介绍了一种在两个单链表中找到它们开始相交节点的方法。通过比较两个链表的长度并调整较长链表的起始位置，使得两链表同时遍历时相遇的第一个节点即为最早公共节点。该算法实现简单且效率高，运行时间为O(n)，仅使用O(1)额外空间。

原文链接：http://blog.youkuaiyun.com/litoupu/article/details/41555557

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:

这里写图片描述
begin to intersect at node c1.

Notes:

1，If the two linked lists have no intersection at all, return null.
2，The linked lists must retain their original structure after the function returns.
3，You may assume there are no cycles anywhere in the entire linked structure.
4，Your code should preferably run in O(n) time and use only O(1) memory.

解

查找两个链表的第一个公共节点，如果两个节点的尾节点相同，肯定存在公共节点
方法：长的链表开始多走（h1的数量 - h2的数量）步，然后和短链表同步往下走，遇到的第一个相同的节点就是最早的公共节点

/**
 * 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;
       ListNode h1=headA;
       ListNode h2=headB;
       int count1=1;
       int count2=1;
       while(h1.next!=null){
           count1++;
           h1=h1.next;
       }

       while(h2.next!=null){
           count2++;
           h2=h2.next;
       }

       if(h1!=h2) return null;
       else{
           int count=Math.abs(count1-count2);
           if(count1>count2){
               h1=headA;
               h2=headB;
           }else{
               h1=headB;
               h2=headA;
           }
           while(count>0){
               h1=h1.next;
               count--;
           }

           while(h1!=null && h2!=null && h1!=h2){
               h1=h1.next;
               h2=h2.next;
           }

           return h1;
       }

      }
}