/*     * 160. Intersection of Two Linked Lists      * 11.28 by Mingyang      * 自己写的代码很长，但是话粗理不粗     */    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {        int lenA = length(headA), lenB = length(headB);        // move headA and headB to the same start point        while (lenA > lenB) {            headA = headA.next;            lenA--;        }        while (lenA < lenB) {            headB = headB.next;            lenB--;        }        // find the intersection until end        while (headA != headB) {            headA = headA.next;            headB = headB.next;        }        return headA;    }    private int length(ListNode node) {        int length = 0;        while (node != null) {            node = node.next;            length++;        }        return length;    }    /*     * 那么下面是更加巧妙的方法，就是直接看两个不相等就继续走     * 知道一个走到null以后回到另一个list的开始继续走     * 第二次停止的时候刚好就是他们相遇的时候     */    public ListNode getIntersectionNode1(ListNode headA, ListNode headB) {        //boundary check        if(headA == null || headB == null) return null;        ListNode a = headA;        ListNode b = headB;        //if a & b have different len, then we will stop the loop after second iteration        while( a != b){            //for the end of first iteration, we just reset the pointer to the head of another linkedlist            a = a == null? headB : a.next;            b = b == null? headA : b.next;            }        return a;    }