Leecode之相交链表

一.题目及剖析

https://leetcode.cn/problems/intersection-of-two-linked-lists/description/

这道题无非就是要做两件事,一是判断链表是否相交,而是找到这个交点

二.思路引入

1.判断链表是否相交只需要判断尾节点地址是否相同(注意一定不能去判断value是否相同)

2.如果尾节点相同,则遍历链表拿到两个链表的长度

3.让长链表先走,走到剩余长度与短链表相同时两者一块走,直到找到一个节点的地址相同,返回该节点

三.代码引入

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
 typedef struct ListNode ListNode;
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    ListNode* curA = headA;
    ListNode* curB = headB;
    int lenA = 0;
    while(curA->next)
    {
        curA = curA->next;
        lenA++;
    }
    int lenB = 0;
    while(curB->next)
    {
        curB = curB->next;
        lenB++;
    }
    if(curA != curB)
    return NULL;
    int gap = abs(lenA - lenB);
    ListNode* longList = headA;
    ListNode* shortList = headB;
    if(lenA < lenB)
    {
        longList = headB;
        shortList = headA;
    }
    while(gap--)
    longList = longList->next;
    while(longList != shortList)
    {
        longList = longList->next;
        shortList = shortList->next;
    }
    return longList;
}