輸入兩個連結串列，找出它們的第一個公共結點。

思路一：因為如果兩個連結串列有公共節點的話，則他們的尾部是一樣的，用兩個棧來存兩個連結串列的節點，開始彈出的值是相同的，彈到不同的時候，就截至，其下一節點就是所求第一個公共節點。

//最快捷的方法
	public static ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
		ListNode p1 = pHead1;
		ListNode p2 = pHead2;
		while(p1!=p2){
			p1 = (p1==null)?pHead2:p1.next;
			p2 = (p2==null)?pHead1:p2.next;
		}
		return p1;
	}
 

package cn.com.jianzhioffer;

import java.util.Stack;

public class Solution37 {
	public static class ListNode {
		int val;
		ListNode next = null;

		ListNode(int val) {
			this.val = val;
		}

		public boolean equals(ListNode pNode1, ListNode pNode2) {
			if (pNode1.val == pNode2.val)
				return true;
			return false;
		}
	}

	public static ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
		if (pHead1 == null || pHead2 == null)
			return null;
		Stack<ListNode> stack1 = new Stack<ListNode>();
		Stack<ListNode> stack2 = new Stack<ListNode>();
		ListNode pNode1 = pHead1;
		ListNode pNode2 = pHead2;
		while (pNode1 != null) {
			stack1.push(pNode1);
			pNode1 = pNode1.next;
		}
		while (pNode2 != null) {
			stack2.push(pNode2);
			pNode2 = pNode2.next;
		}
		pNode1 = stack1.pop();
		pNode2 = stack2.pop();
		while (pNode1.equals(pNode2) && (!stack1.empty()) && (!stack2.empty())) {
			pNode1 = stack1.pop();
			pNode2 = stack2.pop();
		}
		if (stack1.empty() || stack2.empty()) {
			return pNode1;
		} else {
			return pNode1.next;
		}

	}

	public static void main(String[] args) {
		ListNode pHead1 = new ListNode(1);
		ListNode pNode2 = new ListNode(2);
		ListNode pNode3 = new ListNode(3);

		pHead1.next = pNode2;
		pNode2.next = pNode3;

		ListNode pHead2 = new ListNode(4);
		pHead2.next = pNode2;

		System.out.println(FindFirstCommonNode(pHead1, pHead2).val);

	}

}