Leetcode NO.160 Intersection of Two Linked Lists

本题题目要求如下：

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.

本题我开始也想复杂了，写了一大堆链表操作。。但是，经过一段时间的挣扎后，我认为easy的题目不该是这种难度，然后，换了一个角度思考，果然很简单，就拿上面的代码举例子把，我们找到c1的最大障碍就是A和B到c1的长度不一样。。。于是我就想办法让他们一样，最开始我的想法是倒着数:c3,c2,c1。但是感觉不是很好写，于是就换了思路：

想办法让B从b2开始，A从a1开始，检查第一个相同的节点：

具体操作是：得到A和B的长度，比如这里A是5,B是6，则此时不计算B的第一个节点b1，直接从b2开始数

过程：

a1 != b2

a2 != b3

c1 == c1

成功！！

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) {        int len_A = 0;        ListNode* tmp = headA;        while (tmp != NULL) {            tmp = tmp->next;            ++len_A;        }        int len_B = 0;        tmp = headB;        while (tmp != NULL) {            tmp = tmp->next;            ++len_B;        }        int diff = abs(len_A - len_B);        if (len_A > len_B) {            while (diff-- > 0)                headA = headA->next;        }        else if (len_A < len_B) {            while (diff-- > 0)                headB = headB->next;        }        while (headA != NULL) {            if (headA == headB)                return headA;            headA = headA->next;            headB = headB->next;        }        return NULL;    }};