Leetcode——相交链表

方法：

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

思路：

如果两个链表相交，那么从交点之后它们的节点都相同。如果想求出交点，可以先遍历两个链表求出它们的长度。求出长度差。然后创建两个指针分别指向两个链表的头节点，之后让长链表先走长度差步，此时两个指针同时走，并且每走一步都要比较指向的节点是否是同一个节点，如果是，则返回节点，遍历完之后则返回空。