Intersection of Two Linked Lists (求两个单链表的相交结点)

题目描述：

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

solution：

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    if (headA == NULL || headB == NULL)
        return NULL;
    int lenA = 0;
    int lenB = 0;
    ListNode *pA = headA;
    while (pA != NULL)
    {
        ++lenA;
        pA = pA->next;
    }
    ListNode *pB = headB;
    while (pB != NULL)
    {
        ++lenB;
        pB = pB->next;
    }

    if (pA != pB)
        return NULL;

    pA = headA;
    pB = headB;
    
    if (lenA > lenB)
    {
        int k = lenA - lenB;
        while (k--)
        {
            pA = pA->next;
        }
    }
    else
    {
        int k = lenB - lenA;
        while (k--)
        {
            pB = pB->next;
        }
    }
    
    while (pA != pB)
    {
        pA = pA->next;
        pB = pB->next;
    }
    return pA;
}

 　　上述解法来自《剑指offer》，还有一种基于Hashset的方法。

solution：

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    if (headA == NULL || headB == NULL)
        return NULL;
    unordered_set<ListNode*> st;
    while (headA != NULL)
    {
        st.insert(headA);
        headA = headA->next;
    }
    
    while (headB != NULL)
    {
        if (st.find(headB) != st.end())
            return headB;
        headB = headB->next;
    }
    return NULL;
}