#380 Intersection of Two Linked Lists

最新推荐文章于 2024-06-24 11:25:34 发布

原创最新推荐文章于 2024-06-24 11:25:34 发布 · 147 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#lintcode #linked list

lintcode 专栏收录该内容

217 篇文章

订阅专栏

本文介绍了一种在两个单链表中找到它们开始相交节点的方法。通过首先确定两个链表的长度差，使较长链表先移动相应步数来对齐起点，随后同步移动两个链表直至找到相交节点。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

Write a program to find the node at which the intersection of two singly linked lists begins.

Notice

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.

Have you met this question in a real interview?

Yes

Example

The following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

Challenge

Your code should preferably run in O(n) time and use only O(1) memory.

题目思路：

这题我觉得有歧义啊，题目说两个list有intersection，我就不由自主的去考虑他们intersect完了还会分开的可能性。结果想了半天，看了答案才知道题目其实说的是有共同的tail，然后找这个common tail的起始点。这就很简单了：因为tail是相同的，而A和B可能长度不同，所以我们可以先align他们的header，也就是说让长的list先从头走几步，使得新的header到tail的距离和短的list相等。这样，再让两个header同时往前走，如果看到common node就返回。

Mycode（AC = 31ms）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /**
     * @param headA: the first list
     * @param headB: the second list
     * @return: a ListNode
     */
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        // write your code here
        // get length of list A and B
        int lenA = getListLength(headA);
        int lenB = getListLength(headB);
        
        // always assume B is longer than A
        if (lenA > lenB) {
            return getIntersectionNode(headB, headA);
        }
        else {
            int diff = lenB - lenA;
            ListNode *nodeA = headA, *nodeB = headB;
            
            // align B and A's header
            while (diff > 0) {
                nodeB = nodeB->next;
                diff--;
            }
            
            // go ahead together, to find whether
            // the intersection begins
            while (nodeA && nodeB) {
                if (nodeA->val == nodeB->val) {
                    break;
                }
                nodeA = nodeA->next;
                nodeB = nodeB->next;
            }
            
            return nodeA;
        }
    }
    
    int getListLength(ListNode *head) {
        int len = 0;
        ListNode *node = head;
        
        while (node) {
            node = node->next;
            len++;
        }
        
        return len;
    }
};