[leetcode]160. Intersection of Two Linked Lists两链表交点

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). 
From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. 
There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

题意: 

描述：给定两个链表，可能相交于某节点。请找出此节点。

Solution1:  

1.  Find the lengths of two lists

2. Ask longer length list to move diff steps (for loop)

3. Ask two lists to move steps until headA == headB(while loop)

code

/**
 * 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) {
        if (headA == null || headB == null) return null;
        int lenA = getLength(headA);
        int lenB = getLength(headB);
        if (lenA > lenB) {
            for (int i = 0; i < lenA - lenB; ++i) {
                headA = headA.next;
            }
        } else {
            for (int i = 0; i < lenB - lenA; ++i) {
                headB = headB.next;
            }
        }
        while (headA != null && headB != null && headA != headB) {
            headA = headA.next;
            headB = headB.next;
        }
        return (headA != null && headB != null) ? headB : null;
    }

    public int getLength(ListNode head) {
        int count = 0;
        while (head != null) {
            ++count;
            head = head.next;
        }
        return count;
    }
}