Java-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.

查找两个链表的交叉点 即第一次出现相同节点的地方 使用哈希表 时间复杂度满足要求 但空间复杂度太高 代码如下：

Map<ListNode,Integer> hma=new HashMap<ListNode,Integer>();
        Map<ListNode,Integer> hmb=new HashMap<ListNode,Integer>();
        while(headA!=null||headB!=null){
            if(headA!=null){
                if(hmb.containsKey(headA)){
                    return headA;
                }else{
                    hma.put(headA,1);
                    headA=headA.next;
                }
            }
            if(headB!=null){
                if(hma.containsKey(headB)){
                    return headB;
                }else{
                    hmb.put(headB,1);
                    headB=headB.next;
                }
            }
        }
        return null;

第二个思路是 晾着若有公用部分 则链表各自部分的差与整个链表长度的差是相同的 所以 先分别算出链表长度  再让长度长的链表的指针先走 差值步 则保证两指针一起走 能一起到公共部分  此公共部分即时所求解 代码如下：

int lena=0;
        int lenb=0;
        ListNode pa=headA;
        ListNode pb=headB;
        while(pa!=null){
            lena++;
            pa=pa.next;
        }
        while(pb!=null){
            lenb++;
            pb=pb.next;
        }

        if(lena<=lenb){
          int    n=lenb-lena;
            while(n>0){
                headB=headB.next;
                n--;
            }
        }else{
           int  n=lena-lenb;
            while(n>0){
                headA=headA.next;
                n--;
            }
        }
        while(headA!=headB){
            headA=headA.next;
            headB=headB.next;
        }
        return headB;