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.

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

题目要求取两个链表的交点，而且时间复杂度必须是O(n)，所以就不能用嵌套循环的方法。用了如下方法，先计算两个链表的各自长度，将长链表节点向下移动两链表长度差，再计算。

代码

/**
 * 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 Aindex_length=getLength(headA);
        int Bindex_length=getLength(headB);
        int dis=Math.abs(Aindex_length-Bindex_length);

        ListNode Aindex=headA;
        ListNode Bindex=headB;
        if(Aindex_length>=Bindex_length){
          for(int i=0;i<dis;i++){
              Aindex=Aindex.next;
          }
          while(Bindex!=null){
              if(Aindex.val==Bindex.val){return Aindex;}
              else{
                  Aindex=Aindex.next;
                  Bindex=Bindex.next;
                 }
          }
        }

         Aindex=headA;
         Bindex=headB;
      if(Aindex_length<Bindex_length){
          for(int i=0;i<dis;i++){
              Bindex=Bindex.next;
          }
          while(Aindex!=null){
              if(Aindex.val==Bindex.val){return Aindex;}
              else{
                  Aindex=Aindex.next;
                  Bindex=Bindex.next;
                 }
          }
        }

        return null;
    }
    public int getLength(ListNode head){
        ListNode index=head;
        int length=1;
        while(index.next!=null){
            index=index.next;
            length++;
        }
        return length;
    }
}

代码下载：https://github.com/jimenbian/GarvinLeetCode

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