解法1：

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        // 定义两个指针。
        // 获得两个链表的长度，将较长的链表先用指针移动到和短链表一样的长度。
        // 再一个个比较
        ListNode l1 = headA, l2 = headB;
        int len1 = 0, len2 = 0;
        while(headA != null){
            len1++;
            headA = headA.next;
        } 
        while(headB != null){
            len2++;
            headB = headB.next;
        }
        while(len1 > len2){
            l1 = l1.next;
            len1--;
        }
        while(len1 < len2){
            l2 = l2.next;
            len2--;
        }
        while(l1 != l2){
            l1 = l1.next;
            l2 = l2.next;
        }
        return l1;
    }
}

解法2：（自己写法）

结合图片理解一下：

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode l1 = headA, l2 = headB;
        while(l1 != null && l2 != null){
            l1 = l1.next;
            l2 = l2.next;
        }
        if(l1 == null) l1 = headB;
        else l2 = headA;
        while(l1 != null && l2 != null){
           l1 = l1.next;
           l2 = l2.next;
        }
        if(l1 == null) l1 = headB;
        else l2 = headA;
        while(l1 != l2 && l1 != null && l2 != null){
            l1 = l1.next;
            l2 = l2.next;
        }
        return l1;
        // get length of headA and headB,//
    }
}

标答

public ListNode getIntersectionNode(ListNode headA, ListNode headB){
    if(headA == null || headB == null) return null;
    ListNode l1 = headA, l2 = headB;
    while(l1 != l2)(
        l1 = l1 == null ? headB : l1.next;
        l2 = l2 == null ? headA : l2.next;
    )
    return l1;
}