HappyLeetcode42:Intersection of Two Linked Lists

浏览数：25 / 时间：2015年06月11日

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;
            }
        }
    }
};