leetcode 160

160. Intersection of Two Linked Lists

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.

找到两个单链表相交的点。此处假设单链表不存在环。

思路：首先获得两个单链表的长度，得到长度的差值n，然后将指向长链表的指针A先向后移动n位，之后指针A同指向短链表的指针B同时每次移动一位，当A与B指向同一节点时，该节点就是相交的结点。

/**
 * 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;
        }
        ListNode * a = headA;
        ListNode * b = headB;
        int n = 1;
        int m = 1;
        while(headA->next != NULL)
        {
            headA = headA->next;
            n++;
        }
        while(headB->next != NULL)
        {
            headB = headB->next;
            m++;
        }
        if(headA != headB)
        {
            return NULL;
        }
        if(n > m)
        {
            for(int i = 0; i < n-m; i++)
            {
                a = a->next;
            }
            while(a)
            {
                if(a == b)
                {
                    return a;
                }
                else
                {
                    a = a->next;
                    b = b->next;
                }
            }
        }
        else
        {
            for(int i = 0; i < m-n; i++)
            {
                b = b->next;
            }
            while(b)
            {
                if(a == b)
                {
                    return a;
                }
                else
                {
                    a = a->next;
                    b = b->next;
                }
            }
        }
        return NULL;
    }
};

拓展：

http://www.cppblog.com/humanchao/archive/2015/03/22/47357.html