一、题目内容

二、题解

解法1：

思路

代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode A = headA, B = headB;
        while (A != B) {
            A = A != null ? A.next : headB;
            B = B != null ? B.next : headA;
        }
        return A;
    }
}

解法2：

思路

• 先对两条链表进行遍历，分别得到两条链表的长度，并计算差值 d。

• 让长度较长的链表先走 d 步，然后两条链表同时走，第一个相同的节点即是节点。

  代码

  ```java /**

  • Definition for singly-linked list.
  • public class ListNode {
  • int val;
  • ListNode next;
  • ListNode(int x) {
  • val = x;
  • next = null;
  • }

  • } */ public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) {

    if (headA == null || headB == null) {
        return null;
    }
    int lena = 0, lenb = 0;
    ListNode a = headA, b = headB;
    while (a != null) {
        lena++;
        a = a.next;
    }
    while (b != null) {
        lenb++;
        b = b.next;
    }
    int d = lena - lenb;
    if (d > 0) {
        while (d-- > 0) {
            headA = headA.next;
        }
    } else {
        d = -d;
        while (d-- > 0) {
            headB = headB.next;
        }
    }
    while (headA != headB) {
        headA = headA.next;
        headB = headB.next;
    }
    return headA;

    } }

```