链表性质判断

判断单链表是否有环，如果有环的话请返回从头结点进入环的第一个节点，没有就返回null
https://leetcode.com/problems/linked-list-cycle-ii/
两个游标指针一起从头走，一个走一步，一个走两步，若能再次相遇，就说明有环.
两个指针分别从头和相遇处走，再次相遇处就是相交点。

public ListNode detectCycle(ListNode head) {
    ListNode oneStep = head, twoStep = head;
    boolean flag = true;
    while(flag || oneStep != twoStep){
        flag = false;
        if(oneStep != null) oneStep = oneStep.next;
        else return null;
        if(twoStep != null && twoStep.next != null) twoStep = twoStep.next.next;
        else return null;
    }
    for(oneStep = head; oneStep != twoStep; oneStep = oneStep.next, twoStep = twoStep.next);
    return oneStep;
}

判断两个链表是否相交，如果相交，找出交点
https://leetcode.com/problems/intersection-of-two-linked-lists/
从两个链表中长的那个的长度差值处开始一起走，相遇处为交点

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    int lenA = 0, lenB = 0; 
    for(ListNode iter = headA; iter != null; lenA++, iter = iter.next);
    for(ListNode iter = headB; iter != null; lenB++, iter = iter.next);
    int diff = lenA > lenB ? lenA-lenB : lenB-lenA;
    ListNode iterfir = lenA > lenB ? headA : headB;
    ListNode itersec = iterfir == headB ? headA : headB;
    for(int i = 0; i < diff; i++, iterfir = iterfir.next);
    while(iterfir != itersec){
        iterfir = iterfir.next;
        itersec = itersec.next;
    }
    return iterfir;
}