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 = ;

    int lenB = ;

    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;

}

巴特西

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

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