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
                   �K
                     c1 → c2 → c3
                   �J            
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.

 

可以将A,B两个链表看做两部分，交叉前与交叉后。

交叉后的长度是一样的，因此交叉前的长度差即为总长度差。

只要去除这些长度差，距离交叉点就等距了。

为了节省计算，在计算链表长度的时候，顺便比较一下两个链表的尾节点是否一样，

若不一样，则不可能相交，直接可以返回NULL

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * }; */
class Solution 
{public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) 
    {        if(headA == NULL || headB == NULL)            return NULL;            
        int lenA = 1;
        ListNode* curA = headA;        while(curA->next != NULL)
        {
            lenA ++;
            curA = curA->next;
        }        //now curA is the tail of A
        
        int lenB = 1;
        ListNode* curB = headB;        while(curB->next != NULL)
        {
            lenB ++;
            curB = curB->next;
        }        //now curB is the tail of B
        
        if(curA != curB)    //no intersection
            return NULL;        else
        {            //align
            int diff = lenA-lenB;            if(diff > 0)
            {//A go
                while(diff)
                {
                    headA = headA->next;
                    diff--;
                }
            }            else
            {//B go
                while(diff)
                {
                    headB = headB->next;
                    diff++;
                }
            }            //go together
            while(true)
            {//A and B has judged to intersect, no need to avoid NULL
                if(headA == headB)                    return headA;                else
                {
                    headA = headA->next;
                    headB = headB->next;
                }
            }
        }
    }
};