leetCode 160. Intersection of Two Linked Lists 链表

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.

题目大意：

找出两个链表后半部分的交汇点。

思路：

1.求出两个链表的长度。

2.获取链表长度差n。

3.将长的链表先移动到第n个节点。

4.对长链表和短链表进行比较。(同时向后移动)如果在链表尾之前找到相等的节点，返回该节点，如果没找到，返回NULL。

代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    int listLength(ListNode *head)//用快指针求链表长度
    {
        ListNode * p = head;
        int i = 0 ;
        while(p && p->next)
        {
            i++;
            p = p->next->next;
        }
        if(p == NULL)
            return 2 * i;
        return 2 * i + 1;
    }
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        int lenA = listLength(headA);
        int lenB = listLength(headB);
        
        int maxLen = lenA > lenB ? lenA :lenB;
        int remain ; 
        ListNode * la,*lb;
        la = headA;
        lb = headB;
        
        if(maxLen == lenA)
        {
            remain = lenA - lenB;
            while(remain--)
            {
                la = la->next;
            }
        }
        else
        {
            remain = lenB - lenA;
            while(remain--)
                lb = lb->next;
        }
        
        while(lb != NULL)
        {
            if(la != lb)
            {
                la = la->next;
                lb = lb->next;
            }
            else
            {
                return la;
            }
        }
        return NULL;
    }
};

2016-08-13 01:08:14