LeetCode-Intersection of Two Linked Lists

Intersection of Two Linked Lists

##题目

####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.

####Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

##解题思路
这道题是找到两个链表的共同的第一个节点。方法比较简单，首先对两个链表分别遍历一次，分别得到两个链表的长度为m,n,且m<n.然后定义两个指针p,q.p指向长度为m的链表，q指向长度为n的链表，然后首先让q先向前遍历n-m个位置，然后p和q同时一起前进，当q==p时，此时的节点即为两个链表第一个公共节点。

##算法代码
代码采用JAVA实现：

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/**
 * 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;
        //获取两个链表的长度
        ListNode p=headA;
        int headAlen=0;
        int headBlen=0;
        while(p!=null)
        {
        	headAlen++;
        	p=p.next;
        }
        p=headB;
        while(p!=null)
        {
        	headBlen++;
        	p=p.next;
        }
        ListNode longP;
        ListNode shortP;
        if(headAlen<headBlen)
        {
        	longP=headB;
        	shortP=headA;
        }else
        {
        	longP=headA;
        	shortP=headB;
        }

        //longP向前历|headAlen-headBlen|
        int m=Math.abs(headAlen-headBlen);
        while(m!=0)
        {
        	longP=longP.next;
        	m--;
        }
        while(longP!=shortP)
        {
        	longP=longP.next;
        	shortP=shortP.next;
        }
        return longP;
    }
}

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