[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");}