[leedcode 160] 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.

  /**
   * 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) {
          //注意，求长度时，可以利用尾节点是否相等进行剪枝,因此需要注意尾节点不能为null
          
          if(headA==null||headB==null) return null;
          int lenA=1;
          ListNode pA=headA;
          while(pA.next!=null){//判断条件，需要利用尾节点
              lenA++;
              pA=pA.next;
          }
          int lenB=1;
          ListNode pB=headB;
          while(pB.next!=null){
              lenB++;
              pB=pB.next;
          }
          if(pA!=pB) return null;////
          pA=headA;////
          pB=headB;
          if(lenA>lenB){
              int step=lenA-lenB;
              for(;step>0;step--){
                  pA=pA.next;
              }
          }
          if(lenB>lenA){
              int step=lenB-lenA;
              for(;step>0;step--){
                  pB=pB.next;
              }
          }
          while(pA!=null){
              if(pA==pB) return pA;
              pA=pA.next;
              pB=pB.next;
          }
          return null;
      }
  }