leetCode-160:Intersection of Two Linked Lists

问题描述

给定两个链表，要求判断两个链表是否相交，如果相交，则返回相交的节点，否则返回 null。题目链接：点我

样例输入输出

输入：intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3

输出：8

解释：两个链表如下所示

输入：intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1

输出：2

解释：两个链表如下所示

问题解法

先分别计算两个链表的长度，然后将链表长度对齐，从对齐节点开始依次向后遍历链表直到两个指针指向的节点相同（同个节点或者都是null），代码如下

/**
 * 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) {
        int lengthA = calcLength(headA);
        int lengthB = calcLength(headB);
        ListNode pA = headA;
        ListNode pB = headB;
        while (lengthA > lengthB) {
            pA = pA.next;
            lengthA--;
        }
        
        while (lengthB > lengthA) {
            pB = pB.next;
            lengthB--;
        }
        
        while (pA != pB) {
            pA = pA.next;
            pB = pB.next;
        }
        
        return pA;
    }
    
    private int calcLength(ListNode head) {
        int length = 0;
        ListNode p = head;
        while (p != null) {
            length++;
            p = p.next;
        }
        
        return length;
    }
}