题解 | 两个链表的第一个公共结点

链表节点的结构：

/*
public class ListNode {
    int val;
    ListNode next = null;

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

解法一、栈

import java.util.*;
public class Solution {
    public ListNode FindFirstCommonNode(ListNode h1, ListNode h2) {
        ListNode ret = null;
        if (h1 == null || h2 == null) return null;
        Stack<ListNode> stack1 = new Stack<>();
        Stack<ListNode> stack2 = new Stack<>();
        while (h1!=null) {
            stack1.push(h1);
            h1 = h1.next;
        }
        while (h2!=null) {
            stack2.push(h2);
            h2 = h2.next;
        }
        while (!stack1.isEmpty() && !stack2.isEmpty()) {
            if (stack1.peek().val == stack2.peek().val) {
                ret = stack1.peek();
                stack1.pop();
                stack2.pop();
            } else {
                return ret;
            }
        }
        return ret;
    }
}

解法二、将两个链表连到一起

import java.util.*;
public class Solution {
    public 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;
    }
}