力扣刷题记录11.1-----面试题 02.07. 链表相交

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一、题目

二、代码

/**
 * 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) {
        

    ///必定是尾部相交  这个给的比较模糊

    ListNode* count_node = new ListNode(0); 

    ListNode* A_node = new ListNode(0); 
    ListNode* B_node = new ListNode(0);  

    ListNode* return_node = new ListNode(0);  
    return_node=nullptr;

    A_node=headA;
    B_node=headB;

    int length_of_A=0;
    int length_of_B=0;

    count_node=headA;
    while(count_node!=nullptr) 
    {
      length_of_A+=1;
      count_node=count_node->next;
    }
    std::cout<<" length_of_A "<< length_of_A <<std::endl;

    count_node=headB;
    while(count_node!=nullptr) 
    {
      length_of_B+=1;
      count_node=count_node->next;
    }
    std::cout<<" length_of_B "<< length_of_B <<std::endl;

    int difference_of_length=abs(length_of_B-length_of_A);  //计算长度差值
    std::cout<<" difference_of_length "<< difference_of_length <<std::endl;
    
   //将链表的节点统一
    int count_decrease_difference=0;
    if(length_of_A<length_of_B)
    {
      while(count_decrease_difference<difference_of_length)
       {
        count_decrease_difference=count_decrease_difference+1;
        B_node=B_node->next;
       }
    }
    if(length_of_A>length_of_B)
    {
      while(count_decrease_difference<difference_of_length)
       {
        count_decrease_difference=count_decrease_difference+1;
        A_node=A_node->next;
       }
    }
     
     
     while(A_node!=nullptr)
     {
       std::cout<<" 此时节点A的数值   "<< A_node->val <<std::endl;
       std::cout<<" 此时节点B的数值   "<< B_node->val <<std::endl;
       if(A_node->val==B_node->val)
       {
           return_node=A_node;
           break;
       }
       A_node=A_node->next;
       B_node=B_node->next;
     }

     return return_node;              
    }
};

三、运行结果

判题系统有一些问题，此题运行通过大部分即可。