leetcode_Intersection of Two Linked Lists_easy_主要是方法

最新推荐文章于 2025-03-18 05:59:11 发布

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最新推荐文章于 2025-03-18 05:59:11 发布

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分类专栏：数据结构和算法文章标签： leetcode 链表

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本文介绍了一种高效的方法来找到两个单链表相交的节点，通过计算链表长度差并使较长链表先行移动，实现同步遍历找出相交节点。

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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.

方法：计算两个链表的长度差len,然后让长的先走len长度，后面2个链表同步走找重合的节点即可。

/**
 * 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) {
        int lenA=0,lenB=0;
        ListNode *tmpA=headA,*tmpB=headB;
        while(tmpA!=NULL)
        {
            lenA++;
            tmpA=tmpA->next;
        }
        while(tmpB!=NULL)
        {
            lenB++;
            tmpB=tmpB->next;
        }
        int len;
        len=lenA>=lenB?(lenA-lenB):(lenB-lenA);
        ListNode *h1,*h2;
        h1=lenA>=lenB?headA:headB;//长的链表
        h2=lenA<lenB?headA:headB;//短的链表
        while(len-->0)
            h1=h1->next;
        while(h1!=h2)
        {
            h1=h1->next;
            h2=h2->next;
        }
        if(h1!=NULL)
            return h1;
        else
            return NULL;
    }
};