[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。

/*************************************************************

 * @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");

}

LeetCode 328. Odd Even Linked List 给定一个单链表，把所有的奇数节点和偶数节点分别排在一起。请注意，这里的奇数节点和偶数节点指的是节点编号的奇偶性，而不是节点的值的奇偶性。
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