🌗 面试题 02.07. 链表相交

难度： 🌗

问题描述

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解法

/**
 * 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) {
        // 思路：
        // 爱心 - 
        if(headA == null || headB == null) {
            return null;
        }
        ListNode aa = headA;
        ListNode bb = headB;
        while(aa != bb) {
            if(aa != null) {
                aa = aa.next;
            } else {
                aa = headB;
            }
            if(bb != null) {
                bb = bb.next;
            } else {
                bb = headA;
            }
        }
        return aa;
    }
}

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输出