[LeetCode] 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:

begin to intersect at node c1.

• 解题思路
  首先把两个链表的所有非空结点入栈，然后比较栈顶元素并出栈，直到一个栈为空或者栈顶元素不相等，此时上一次比较的结点即为相交结点。
• 实现代码

/*************************************************************
    *  @Author   : 楚兴
    *  @Date     : 2015/2/8 13:33
    *  @Status   : Accepted
    *  @Runtime  : 76 ms
*************************************************************/
#include <vector>
#include <iostream>
#include <stack>
using namespace std;

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        stack<ListNode*> stack1;
        stack<ListNode*> stack2;
        ListNode* pa = headA;
        ListNode* pb = headB;
        while(pa)
        {
            stack1.push(pa);
            pa = pa->next;
        }
        while(pb)
        {
            stack2.push(pb);
            pb = pb->next;
        }

        ListNode* temp = NULL;
        while (!stack1.empty() && !stack2.empty())
        {
            if (stack1.top()->val == stack2.top()->val)
            {
                temp = stack1.top();
                stack1.pop();
                stack2.pop();
            }
            else
            {
                return temp;
            }
        }
        return temp;
    }
};

void addListNode(ListNode*& head, int i)
{
    ListNode* h = head;
    while(h->next != NULL)
    {
        h = h->next;
    }
    ListNode* temp = new ListNode(i);
    h->next = temp;
}

//void print(ListNode* head)
//{
//  ListNode* temp = head;
//  while(temp)
//  {
//      cout<<temp->val<<endl;
//      temp = temp->next;
//  }
//}

int main()
{
    Solution s;
    ListNode* headA = new ListNode(0);
    addListNode(headA, 1);
    addListNode(headA, 2);
    addListNode(headA, 1);
    addListNode(headA, 2);
    addListNode(headA, 3);

    ListNode* headB = new ListNode(0);
    addListNode(headB, 1);
    addListNode(headB, 2);
    addListNode(headB, 3);
    addListNode(headB, 1);
    addListNode(headB, 2);
    addListNode(headB, 3);

    ListNode* p = s.getIntersectionNode(headA, headB);
    cout<<p->val<<endl;
    system("pause");
}