160. Intersection of Two Linked Lists

    /*
     * 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;
    }