160. Intersection of Two Linked Lists 找两个链表相交的交点

160. Intersection of Two Linked Lists

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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.
 * 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 len1 = 1;
        ListNode *TailA = headA;
        while(TailA->next != NULL)
        {
            TailA = TailA->next;
            len1++;
        }
        
        int len2 = 1;
        ListNode *TailB = headB;
        while(TailB->next != NULL)
        {
            TailB = TailB->next;
            len2++;
        }
        
        if(TailA != TailB) return NULL;
        
        ListNode *NodeA = headA;
        ListNode *NodeB = headB;
        if(len1 > len2)
        {
            int k = len1 - len2;
            while(k--)
                NodeA = NodeA->next;
        }
        else
        {
            int k = len2 - len1;
            while(k--)
                NodeB = NodeB->next;
        }
        
        while(NodeA != NodeB)
        {
        NodeA = NodeA->next;
        NodeB = NodeB->next;
        }
        return NodeA;
    }
};