【LeetCode】25.Linked List —Intersection of Two Linked Lists 链表-两个链表的交集

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

For example, the following two linked lists:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node‘s value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node‘s value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

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.

首先定义两个指针，分别遍历两个链表，得到链表长度。然后求长度差，得出两个链表的差值。开头定义的两个指针回到链表头部，让较长链表的指针先移动差值距离，然后两个指针一起向后移动，直到两个指针所指向的地址相同，即为所求交叉点。

/**
 * 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 *A =headA;
        ListNode *B =headB;
        int countA=0;
        int countB=0;
        int difference;
        while(A!=NULL)
        {
            A=A->next;
            countA++;
        }
        while(B!=NULL)
        {
            B=B->next;
            countB++;
        }
        A=headA;
        B=headB;
        if(countA>=countB)
        {
           difference=countA-countB;
           for(int i=0;A!=NULL&&i<difference;i++)
           {
               A=A->next;
           }
           while(A!=NULL&&B!=NULL)
           {
               if(A==B) return A;
               else
               {
                   A=A->next;
                   B=B->next;
               }
           }
           return NULL;
        }
        else
        {
            difference=countB-countA;
            for(int i=0;B!=NULL&&i<difference;i++)
            {
                B=B->next;
            }
            while(B!=NULL&&A!=NULL)
            {
                if(A==B)return B;
                else
                {
                    A=A->next;
                    B=B->next;
                }
            }
            return NULL;
        }
        
    }
};

【LeetCode】25.Linked List —Intersection of Two Linked Lists 链表-两个链表的交集

原文：https://www.cnblogs.com/hu-19941213/p/11255446.html