题解 | #两个链表的第一个公共结点#

public class Solution {
    //计算链表长度的函数 fast-template
    public int ListLenth(ListNode pHead){
        ListNode p = pHead;
        int n = 0;
        while(p != null){
            n++;
            p = p.next;
        }
        return n;
    }
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
        int p1 = ListLenth(pHead1);
        int p2 = ListLenth(pHead2);
        //当链表1更长时，链表1指针先走p1-p2步
        if(p1 >= p2){
            int n = p1 - p2;
            for(int i = 0; i < n; i++){
                pHead1 = pHead1.next;
            }
            //两个链表同时移动，直到有公共结点时停下
            while((pHead1 != null) && (pHead2 != null) && (pHead1 != pHead2)){
                pHead1 = pHead1.next;
                pHead2 = pHead2.next;
            }
        }
        //反之，则链表2先行p2-p1步
        else{
            int n = p2 - p1;
            for(int i = 0; i < n; i++){
                pHead2 = pHead2.next;
            }
            //两个链表同时移动，直到有公共结点时停下
            while((pHead1 != null) && (pHead2 != null) && (pHead1 != pHead2)){
                pHead1 = pHead1.next;
                pHead2 = pHead2.next;
            }
        }
        return pHead1;
    }
}