链表相交问题

链接
判断两个链表是否相交

分为如下几个问题：
1.判断链表是否有环，若有返回入环节点
2.两个无环链表判断相交
3.两个有环链表判断相交
4.一个无环链表与一个有环链表必不相交

public class Solution {

    /**
     * 判断链表是否有环
     *
     * @param head
     * @return 如果有环，返回入环节点，否则，返回null
     */
    private static ListNode hasLoop(ListNode head) {
        if (head == null || head.next == null || head.next.next == null) {
            return null;
        }
        ListNode slow = head.next, fast = head.next.next;

        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                slow = head;
                while (slow != fast) {
                    slow = slow.next;
                    fast = fast.next;
                }
                return slow;
            }
        }
        return null;
    }

    /**
     * 判断两个无环链表是否相交
     *
     * @param head1
     * @param head2
     * @return 第一个相交节点
     */
    private static ListNode judgeTwoNoLoopList(ListNode head1, ListNode head2) {
        if (head1 == null || head2 == null) {
            return null;
        }

        int n = 0;

        ListNode cur1 = head1, tail1 = head1;
        while (cur1 != null) {
            n++;
            tail1 = cur1;
            cur1 = cur1.next;
        }

        ListNode cur2 = head2, tail2 = head2;
        while (cur2 != null) {
            n--;
            tail2 = cur2;
            cur2 = cur2.next;
        }

        if (tail1 != tail2) {
            return null;
        }

        cur1 = n > 0 ? head1 : head2;
        cur2 = cur1 == head1 ? head2 : head1;

        n = Math.abs(n);

        while (n-- > 0) {
            cur1 = cur1.next;
        }

        while (cur1 != null && cur2 != null) {
            if (cur1 == cur2) {
                return cur1;
            }
            cur1 = cur1.next;
            cur2 = cur2.next;
        }

        return null;
    }


    /**
     * 判断两个有环链表是否相交
     *
     * @param head1
     * @param head2
     * @param loop1 入环节点
     * @param loop2 入环节点
     * @return 第一个相交节点
     */
    private static ListNode judgeTwoLoopList(ListNode head1, ListNode loop1, ListNode head2, ListNode loop2) {
        if (head1 == null || head2 == null) {
            return null;
        }


        if (loop1 == loop2) {
            return loop1;
        }

        ListNode cur = head1.next;

        while (cur != head1) {
            if (cur == loop2) {
                return cur;
            }
            cur = cur.next;
        }

        return null;
    }

    public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null && headB != null || headA != null && headB == null) {
            return null;
        }

        ListNode loop1 = hasLoop(headA);
        ListNode loop2 = hasLoop(headB);

        if (loop1 == null && loop2 == null) {
            return judgeTwoNoLoopList(headA, headB);
        } else if (loop1 != null && loop2 != null) {
            return judgeTwoLoopList(headA, loop1, headB, loop2);
        }

        return null;
    }
}

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
        next = null;
    }

    @Override
    public String toString() {
        return "ListNode{" +
                "val=" + val +
                '}';
    }
}

标签：问题,ListNode,head1,cur1,相交,next,链表,return,null
来源： https://www.cnblogs.com/tianyiya/p/15402430.html