剑指offer 两个链表的第一个公共结点

题目描述

输入两个链表，找出它们的第一个公共结点。思路：找出2个链表的长度，然后让长的先走两个链表的长度差，然后再一起走（因为2个链表用公共的尾部）/*
public class ListNode {
int val;
ListNode next = null;

ListNode(int val) {
this.val = val;
}
}*/
public class Solution {
public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
int len1 = findListLenth(pHead1);
int len2 = findListLenth(pHead2);
ListNode current1 = pHead1;
ListNode current2 = pHead2;
// 先在长链表上走上几步，在同时在两个链表上遍历
if (len1 - len2 > 0)
current1 = walkStep(current1, len1 - len2);
else
current2 = walkStep(current2, len2 - len1);
while (current1 != null && current2 != null) {
if (current1 == current2)
return current1;
current1 = current1.next;
current2 = current2.next;
}
return null;
}

private ListNode walkStep(ListNode cur, int step) {
// 从step~1
while (step-- > 0)
cur = cur.next;
return cur;
}

// 计算链表长度
private int findListLenth(ListNode head) {
if (head == null)
return 0;
ListNode cur = head;
int length = 0;
while (cur != null) {
length++;
cur = cur.next;
}
return length;
}
}