Intersection of Two Linked Lists 判断两个单链表里边有没有重叠，返回重叠的第一个节点

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./**
• 题目重叠的意思是后边的都得是相同的，光交叉相等一个是不行的(除了最后的节点)，如果是单向链表，两个有公共结点而部分重合的链表，拓扑形状看起来像一个Y，而不可能像X。
•  * 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) {
          int a=0,b=0,sub,i=0;    //必须要初始化，这儿不初始化，下边也得要赋值，不然报错
          ListNode nodeA=headA,nodeB=headB;    //看，可以定义在同一行
          if(headA==null || headB==null)  return null;
          else{
              while(nodeA.next!=null){
                  nodeA=nodeA.next;
                  ++a;
              }
              while(nodeB.next!=null){
                  nodeB=nodeB.next;
                  ++b;
              }
          }
          sub=a>=b? a-b : b-a;   //可以使用Math.abs(a-b)；
          nodeA=a>=b? headA : headB;
          nodeB=a>=b? headB : headA;
  
  
          while(i<sub){
              nodeA=nodeA.next;
              ++i;
          }
          while(nodeA!=nodeB){  //不错哦，如果一旦相等就返回
              nodeA=nodeA.next;
              nodeB=nodeB.next;
          }
         return nodeA;  //包含了相等于不相等两种情况
      }
  }

思路一样，写法稍微有点差别：http://www.1point3acres.com/bbs/thread-110377-1-1.html

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) return null;
        int lenA = length(headA);
        int lenB = length(headB);
        int diff = Math.abs(lenA - lenB);
        while(diff > 0) {
            if (lenA > lenB) headA = headA.next;
            else headB = headB.next;
            diff--;
        }
        while (headA != null && headB != null) {
            if (headA.val == headB.val) return headA;
            headA = headA.next;
            headB = headB.next;
        }
        return null;
    }


    private int length(ListNode n) {
        if (n == null) return 0;
        int length = 0;
        while (n != null) {
            length++;
            n = n.next;
        }
        return length;
    }
    
}