Intersection of Two Linked Lists

存在两个单向链表，如果两者在某个节点相交合并，则返回交点，不相交则返回null。

思路：如果两个链表长度相等，则两个指针依次前进，到结尾之前如果遇到val值相等，则返回该节点。如果不相等，则长的那个链表先走若干步，使得剩下的步数与短的链表的步数相同。

注意点

• 空指针

• 由于是单链表，需要保存链表头

  class Solution {
  public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (headA == NULL || headB == NULL)
        {
            return NULL;
        }
        int lengthA = 0, lengthB = 0;
        ListNode* tmpA = headA;
        ListNode* tmpB = headB;
        while (tmpA->next != NULL)
        {
            lengthA++;
            tmpA = tmpA->next;
        }
        while (tmpB->next != NULL)
        {
            lengthB++;
            tmpB = tmpB->next;
        }
        if (tmpA->val != tmpB->val)
            return NULL;
        tmpA = headA;
        tmpB = headB;
        if (lengthA >lengthB)
        {
            for (size_t i = 0; i < lengthA-lengthB; i++)
            {
                headA = headA->next;
            }
        }
        else
        {
            for (size_t i = 0; i < lengthB - lengthA; i++)
            {
                headB = headB->next;
            }
        }
        while (headA != NULL)
        {
            if (headA->val == headB->val)
            {
                return headA;
            }
            else
            {
                headA = headA->next;
                headB = headB->next;
            }
        }
    }
  };

来自为知笔记(Wiz)