题解 | #两个链表的第一个公共结点#

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

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
 
        ListNode p1 = pHead1;
        ListNode p2 = pHead2;
        //1先遍历一遍 获取链表长度
        int plen1 = 0,plen2 = 0;
        while(p1!=null){
            plen1++;
            p1 = p1.next;
        }
        while(p2!=null){
            plen2++;
            p2 = p2.next;
        }
        //2确定长短链表 ，并创建2个数组存放链表节点
        int maxlen =  plen1>plen2 ? plen1:plen2;
        int minlen =  maxlen==plen2? plen1:plen2;
        
        ListNode[] arr1 = new ListNode[maxlen];
        ListNode[] arr2 = new ListNode[maxlen];
        
        ListNode maxNode = maxlen == plen2? pHead2 : pHead1;
        ListNode minNode = maxNode == pHead1? pHead2 : pHead1;
        
        //存放链表节点
        int i=0;
        while(maxNode!=null){
            arr1[i] = maxNode;
            i++;
            maxNode = maxNode.next;
        }
        i= maxlen- minlen;
        while(minNode!=null){
            arr2[i] = minNode;
            i++;
            minNode = minNode.next;
        }
        //3倒叙 取数组内链表节点，比较值，相同就赋值给输出指针，不同就终止
        ListNode res = null;
        for(int j=maxlen-1;j>=maxlen-minlen;j--){
            if(arr1[j].val == arr2[j].val){
                res = arr1[j];
            }else{
                break;
            }
        }
        return res;
    }
}