• 原题连接：160. 相交链表

1- 思路

思路

• 首先想要找到相交点，需要定义连个指针。两个指针一定得是同步的，例如 A 链表 [1,2,3,4,5] ，链表 B 是 [4,5]
  • 1- 指针对齐：则需要指针1 先移动，移动长度为 lenA - lenB
  • 2- 同时移动：移动指针1 和 指针2 ，遇到相等则返回相等结点，否则返回 null

2- 实现

⭐160. 相交链表——题解思路

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode nodeA = headA;
        ListNode nodeB = headB;
        int lenA = 0;
        int lenB = 0;
        while(nodeA!=null){
            lenA++;
            nodeA = nodeA.next;
        }
        while(nodeB!=null){
            lenB++;
            nodeB = nodeB.next;
        }
        if(lenB>lenA) return getIntersectionNode(headB,headA);
        nodeA = headA;
        nodeB = headB;
        // 1. 先移动第一个指针 
        for(int i = 0 ; i < lenA-lenB;i++){
            headA = headA.next;
        }
        while(headA!=null || headB!=null){
            if(headA==headB){
                return headA;
            }
            headA = headA.next;
            headB = headB.next;
        }
        return null;
    }
}

3- ACM 实现

public class getIntersectionNode {
    public static class ListNode {
        int val;
        ListNode next;
        ListNode(int x) {
            val = x;
            next = null;
        }
    }

    public static ListNode getIntersection(ListNode head1,ListNode head2){
        ListNode curA = head1;
        ListNode curB = head2;
        int len1 = 0;
        int len2 = 0;
        while(curA!=null){
            len1++;
            curA = curA.next;
        }
        while(curB!=null){
            len2++;
            curB = curB.next;
        }
        if(len1<len2) return getIntersection(head2,head1);
        // 遍历
        curA = head1;
        curB = head2;
        while (curA!=null || curB !=null){
            if(curA == curB){
                return curA;
            }
            curA = curA.next;
            curB = curB.next;
        }
        return null;
    }


    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        // 读取第一个链表的节点数量
        int n1 = sc.nextInt();
        ListNode head1 = null, tail1 = null;
        for (int i = 0; i < n1; i++) {
            int val = sc.nextInt();
            ListNode newNode = new ListNode(val);
            if (head1 == null) {
                head1 = newNode;
                tail1 = newNode;
            } else {
                tail1.next = newNode;
                tail1 = newNode;
            }
        }

        // 读取第二个链表的节点数量
        int n2 = sc.nextInt();
        ListNode head2 = null, tail2 = null;
        for (int i = 0; i < n2; i++) {
            int val = sc.nextInt();
            ListNode newNode = new ListNode(val);
            if (head2 == null) {
                head2 = newNode;
                tail2 = newNode;
            } else {
                tail2.next = newNode;
                tail2 = newNode;
            }
        }

        // 读取交点位置
        int intersectionIndex = sc.nextInt();
        if (intersectionIndex >= 0) {
            ListNode intersectionNode = head1;
            for (int i = 0; i < intersectionIndex; i++) {
                if (intersectionNode != null) {
                    intersectionNode = intersectionNode.next;
                }
            }
            if (tail2 != null) {
                tail2.next = intersectionNode;
            }
        }
        System.out.println("结果是"+getIntersection(head1,head2).val);
    }
}