leetcode：Intersection of Two Linked Lists(两个链表的交叉点)

最新推荐文章于 2025-08-15 23:10:13 发布

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最新推荐文章于 2025-08-15 23:10:13 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/carsonzhu/p/4677687.html

本文介绍了如何找到两个单链表开始相交的节点，并提供了多种高效解决方案，包括通过比较链表长度来对齐并同步遍历的方法。

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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.

分析：题意即为寻找两个单链表相交开始的节点

如果我们直接通过遍历两个链表来寻找交叉元点

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:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(headA==NULL || headB==NULL) return NULL;
        ListNode *result=NULL;
        while(headB){
       if(headA && headA->val == headB->val) {
        result = headA; 
       }
        while (headA) 
        { 
            if (headA->next && headA->next->val == headB->val) {
                result=headA->next;
            }
            headA = headA->next; 
        }
        if(headB->next) headB= headB->next;
      }
   
        return result; 
    }
};

会出现超时：Time Limit Exceeded　　

看来还是得认真审题加分析，找到解题的关键特征来提高效率。（一个星期没刷题就变得头脑简单了）

最简单的代码是这样子滴：

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {

        while (headA && headB) {
            if (headA->val < headB->val)
                headA = headA->next;
            else if (headA->val > headB->val)
                headB = headB->next;
            else if (headA->val == headB->val)
                return headA;
        }
        return nullptr;
    }
};

当然也可以是这样的思路：　　

我们可以遍历两个链表得到各自的长度，然后将长度更长的链表向前移动，使两个链表进行对齐，之后一起遍历，直到找到第一个相同的节点。

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {

        auto currA = headA, currB = headB;
        int countA = 0, countB = 0;

        while (currA) { 
            currA = currA->next, countA++;
        }
        while (currB) {
            currB = currB->next, countB++;
        }
        int diff = std::abs(countA - countB);
        if (countB > countA) { swap(headA, headB); }
        while (diff--) { 
            headA = headA->next;
        }
        while (headA != headB) {
            headA = headA->next, headB = headB->next;
        }
        return headA;
    }
};

　或者：

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

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(headA == NULL || headB == NULL)
            return NULL;
        ListNode* iter1 = headA;
        ListNode* iter2 = headB;
        int len1 = 1;
        while(iter1->next != NULL)
        {
            iter1 = iter1->next;
            len1 ++;
        }
        int len2 = 1;
        while(iter2->next != NULL)
        {
            iter2 = iter2->next;
            len2 ++;
        }
        if(iter1 != iter2)
            return NULL;
        if(len1 > len2)
        {
            for(int i = 0; i < len1-len2; i ++)
                headA = headA->next;
        }
        else if(len2 > len1)
        {
            for(int i = 0; i < len2-len1; i ++)
                headB = headB->next;
        }
        while(headA != headB)
        {
            headA = headA->next;
            headB = headB->next;
        }
        return headA;
    }
};

转载于:https://www.cnblogs.com/carsonzhu/p/4677687.html