无环单链表判相交

现在有两个无环单链表，若两个链表的长度分别为m和n，请设计一个时间复杂度为O(n + m)，额外空间复杂度为O(1)的算法，判断这两个链表是否相交。

给定两个链表的头结点headA和headB，请返回一个bool值，代表这两个链表是否相交。保证两个链表长度小于等于500

思路：先判断A和B的长度，A>B的话，就c =A - B，然后判断A 和B 的节点值是否想到直到最后一个，如果在比较的过程中有一个不相等就返回false

代码如下：

package lianbiao;public class danlianbiaoxiangjiao {public static void main(String[] args) {            ListNode A1 =new ListNode(1);            ListNode A2 =new ListNode(2);            ListNode A3 =new ListNode(3);            ListNode A4 =new ListNode(4);            A1.next =A2;            A2.next =A3;            A3.next =A4;            ListNode B1 =new ListNode(2);            ListNode B2 =new ListNode(3);            ListNode B3 =new ListNode(4);            B1.next =B2;            B2.next =B3;            System.out.println(danlianbiaoxiangjiao.chkIntersect(A1, B1));}public static boolean chkIntersect(ListNode headA, ListNode headB) {     ListNode A = headA;     ListNode B = headB;     int countA =0;     int countB =0;     while(A != null){     countA++;     A = A.next;     }     while(B != null){     countB++;     B = B.next;     }          boolean aa =false;     int i = 0;     if(countA > countB){      aa = true;      i =countA - countB;     }else{      i =countB -countA;     }     for(int j=0;j<i;j++){     if(aa){     headA = headA.next;     }else{     headB = headB.next;     }     }     while(headA !=null && headB != null){ if(headA.val != headB.val){  return false; } headA =headA.next; headB =headB.next; } return true;          }static class  ListNode{int val = 0;ListNode next = null;ListNode(int val){this.val = val;}}}