两个链表的第一个公共节点

题目

输入两个链表，找出它们的第一个公共节点。

分析

代码

class ListNode{
    public int data;
    public ListNode next;

    public ListNode(int data) {
        this.data = data;
    }

    @Override
    public String toString() {
        return "ListNode{" +
                "data=" + data +
                '}';
    }
}

import java.util.Stack;

public class Solution {

    public static void main(String[] args) {
        //链表1
        ListNode head = new ListNode(8);
        ListNode node1 = new ListNode(1);
        ListNode node2 = new ListNode(8);
        ListNode node3 = new ListNode(4);
        ListNode node4 = new ListNode(5);
        head.next = node1;
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        //链表2
        ListNode head2 = new ListNode(4);
        ListNode node6 = new ListNode(1);
        ListNode node7 = new ListNode(8);
        head2.next = node6;
        node6.next = node7;
        node7.next = node2;
        System.out.println(getIntersectionNode(head,head2));
    }

    /**
     * 寻找两个链表的第一个公共节点
     * @param headA 链表
     * @param headB 链表
     * @return 第一个公共节点
     */
    public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        Stack<ListNode> stackA = ListNode2Stack(headA);
        Stack<ListNode> stackB = ListNode2Stack(headB);
        ListNode temp = null;
        while (!stackA.isEmpty() && !stackB.isEmpty()){
            if(stackA.peek() == stackB.peek()) {
                stackA.pop();
                temp = stackB.pop();
            }
            else {
                break;
            }
        }
        return temp;
    }

    /**
     * 遍历单向链表 将每个节点压入栈中
     * @param head
     */
    public static Stack<ListNode> ListNode2Stack(ListNode head){
        Stack<ListNode> stack = new Stack<ListNode>();//创建栈
        while (head != null){
            stack.add(head);//将每个节点添加到栈中
            head = head.next;
        }
        return stack;
    }
}

分析

代码

public class Solution {

    public static void main(String[] args) {
        //链表1
        ListNode head = new ListNode(8);
        ListNode node1 = new ListNode(1);
        ListNode node2 = new ListNode(8);
        ListNode node3 = new ListNode(4);
        ListNode node4 = new ListNode(5);
        head.next = node1;
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        //链表2
        ListNode head2 = new ListNode(4);
        ListNode node6 = new ListNode(1);
        ListNode node7 = new ListNode(8);
        head2.next = node6;
        node6.next = node7;
        node7.next = node2;
        System.out.println(getIntersectionNode(head,head2));

    }

    /**
     * 寻找两个链表的第一个公共节点
     * @param headA 链表
     * @param headB 链表
     * @return 第一个公共节点
     */
    public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        int lenA = getListLength(headA);//链表长度
        int lenB = getListLength(headB);
        //链表1长度大于链表2
        if (lenA > lenB){
            //先将链表1向前走 lenA-lenB
            for (int i = 0; i < lenA-lenB; i++){
                headA = headA.next;
            }
        }else {
            for (int i = 0; i < lenB-lenA; i++){
                headB = headB.next;
            }
        }
        for (int i = 0; i < Math.min(lenA,lenB); i++){
            if (headA == headB){
                break;
            }
            headA = headA.next;
            headB = headB.next;
        }
        return headA;
    }

    /**
     * 获取链表长度
     * @param head 头节点
     * @return 链表长度
     */
    public static int getListLength(ListNode head){
        int count = 0;
        while (head != null){
            count++;
            head = head.next;
        }
        return count;
    }

分析

代码

public class Solution {

    public static void main(String[] args) {
        //链表1
        ListNode head = new ListNode(8);
        ListNode node1 = new ListNode(1);
        ListNode node2 = new ListNode(8);
        ListNode node3 = new ListNode(4);
        ListNode node4 = new ListNode(5);
        head.next = node1;
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        //链表2
        ListNode head2 = new ListNode(4);
        ListNode node6 = new ListNode(1);
        ListNode node7 = new ListNode(8);
        head2.next = node6;
        node6.next = node7;
        node7.next = node2;
        System.out.println(getIntersectionNode(head,head2));

    }
    /**
     * 寻找两个链表的第一个公共节点
     * @param headA 链表
     * @param headB 链表
     * @return 第一个公共节点
     */
    public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode node1 = headA;
        ListNode node2 = headB;
        //同时遍历两个链表
        while (node1 != node2){
            node1 =  node1 == null? headB : node1.next;
            node2 =  node2 == null? headA : node2.next;
        }
        return node1;
    }

}