[Leedcode] Necessary interview questions for linked lists in data structures (Phase 4)

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[Leedcode] Necessary interview questions for linked lists in data structures (Phase 4)
Summarize

1. Topic

Intersecting linked list: as follows (example):

给你两个单链表的头节点 headA 和 headB ，请你找出并返回两个单链表相交的起始节点。如果两个链表不存在相交节点，返回 null 。
图示两个链表在节点 c1 开始相交：题目数据 保证 整个链式结构中不存在环。
注意，函数返回结果后，链表必须 保持其原始结构 。

1. Determine whether two linked lists intersect? 2. If it intersects, find the intersection point

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2. Idea + diagram

(1) Idea 1

Brute force solution - exhaustive method. Take each node in the A linked list in turn and compare it with all the nodes in the B linked list.
If there are identical nodes, it is an intersection, and the first identical intersection is a common node. The time complexity of doing this is: O(N^2)
So how do we optimize the time complexity to: O(N)

(2) Idea 2

1. If the end nodes are the same, they intersect, otherwise they will not intersect.
2. Find the intersection point: the long linked list takes (length difference) steps first, and then walks at the same time. The first identical node is the intersection point , as shown in the figure below

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Note here: You can use lenA and lenB to calculate the length of the two linked lists, which is convenient for finding the position of the intersection point , as shown in the figure below
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3. Source code

The code is as follows (example):

struct ListNode 
{
    
    
     int val;
     struct ListNode *next;
};
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB)
{
    
    
    struct ListNode* pheadA = headA;
    struct ListNode* pheadB = headB;
    //先判断是否为环形结构
    int lenA = 1;
    while(pheadA -> next)
    {
    
    
        lenA++;
        pheadA = pheadA -> next;
    }
    int lenB = 1;
    while(pheadB -> next)
    {
    
    
        lenB++;
        pheadB = pheadB -> next;
    }
    if(pheadA != pheadB)
    {
    
    
        return NULL;
    }
    int sub = abs(lenA - lenB);
    struct ListNode* longlist = headA;
    struct ListNode* shortlist = headB;
    if(lenA < lenB)
    {
    
    
        longlist = headB;
        shortlist = headA;
    }
    //长的先走sub步
    while(sub--)
    {
    
    
        longlist = longlist -> next;
    }
    //俩个开始一起走
    while(longlist != shortlist)
    {
    
    
        longlist = longlist -> next;
        shortlist = shortlist -> next;
    }
    return longlist;
}

Summarize

The above is what I want to talk about today. This article introduces the necessary interview questions for linked lists in data structures (Phase 4).
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