HappyLeetcode42:Intersection of Two Linked Lists

最新推荐文章于 2021-06-08 01:31:36 发布

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最新推荐文章于 2021-06-08 01:31:36 发布

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原文链接：http://www.cnblogs.com/chengxuyuanxiaowang/p/4209448.html

本文介绍了一种在两个单链表中找到它们开始相交节点的方法。通过计算两个链表的长度并调整较长链表的起点，确保两者同时到达交点。此方法运行时间为O(n)，仅使用O(1)额外内存。

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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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

这道题除了阅读题目的困难较大之外，其他的都还好。非常好做的一道题目。

代码如下：

//* Definition for singly-linked list.
 struct ListNode {
     int val;
     ListNode *next;
     ListNode(int x) : val(x), next(NULL) {}
 };

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (headA == NULL || headB == NULL)
            return NULL;
        int lengtha=1, lengthb=1;
        ListNode *cora = headA, *corb = headB;
        while (cora->next != NULL)
        {
            cora = cora->next;
            lengtha++;
        }
        while (corb->next != NULL)
        {
            corb = corb->next;
            lengthb++;
        }
        if (cora->val != corb->val)
            return NULL;
        int dif = lengtha - lengthb;
        cora = headA, corb = headB;
        if (dif >= 0)
        {
            while (dif)
            {
                dif--;
                cora = cora->next;
            }
        }
        else

        {
            while (dif)
            {
                dif++;
                corb = corb->next;
            }
        }

        while (cora != NULL)
        {
            if (cora == corb)
                return cora;
            else
            {
                cora = cora->next;
                corb = corb->next;
            }
        }
    }
};