判断两个单向链表是否相交

Posted 2021-05-07

分为三种情况：
第一种情况: 一个链表有环，一个链表没有环，那这两个链表不可能相交
第二种情况: 两个链表都没有环
第三种情况: 两个链表都有环

public class  FindFirstIntersectNode {

    public static class Node {
        public int value;
        public Node next;

        public Node(int data) {
            this.value = data;
        }
    }

    /**
     *  分为三种情况：
     *  第一种情况: 一个链表有环，一个链表没有环，那这两个链表不可能相交。
     *  第二种情况: 两个链表都没有环，
     *  第三种情况: 两个链表都有环，
     */
    public static Node getIntersectNode(Node head1, Node head2) {
        if (head1 == null || head2 == null) {
            return null;
        }
        Node loop1 = getLoopNode(head1);
        Node loop2 = getLoopNode(head2);
        
        //两个无环链表是否相交
        if (loop1 == null && loop2 == null) {
            return noLoop(head1, head2);
        }
        
        //两个有环链表是否相交
        if (loop1 != null && loop2 != null) {
            return bothLoop(head1, loop1, head2, loop2);
        }
        return null;
    }

    
    /**
     * 判断链表是否有环，链表有环返回第一个入环的节点，没有环返回null
     * @param head
     * @return
     */
    public static Node getLoopNode(Node head) {
       //先求出快慢指针的相遇点cross
        Node fast = head;
        Node slow = head;
        Node cross = null;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (slow == fast) {
                cross = fast;
                break;
            }
        }
        //从链表头部和相遇点开始，每次移动一个节点,他们相遇点就是环的入口
        Node start = head;
        while (start != null && cross != null) {
            if (start == cross) {
                return cross;
            }
            start = start.next;
            cross = cross.next;
        }
        return null;
    }

    
    /**
     * 两个无环链表是否相交，相交返回第一个相交节点，不相交返回null
     * @param head1
     * @param head2
     * @return
     */
    public static Node noLoop(Node head1, Node head2) {
        if (head1 == null || head2 == null) {
            return null;
        }
        Node cur1 = head1;
        int  len1 = 1;
        Node cur2 = head2;
        int  len2 = 1;
        while(cur1.next != null) {
            len1++;
            cur1 = cur1.next;
        }
        while(cur2.next != null) {
            len2++;
            cur2 = cur2.next;
        }
        //最后一个节点不相同，两个链表不相交
        if(cur1 != cur2) {
            return null;
        }
        //first 始终指向长链表的开始，second始终指向短链表的开始,diff表示两个链表的长度差
        Node first = null;
        Node second = null;
        int diff = 0;
        if(len1 > len2) {//链表1比链表2要长，
            first = head1;
            second = head2;
            diff = len1 -len2;
        }else {
            first = head2;
            second = head1;
            diff = len2 - len1;
        }
        //在diff不为零的前提下，长链表先走diff
        while(diff > 0) {
            first = first.next;
            diff--;
        }
        //first和second同时走,如果相同了，表示相加了
        while(first != second) {
            first =first.next;
            second = second.next;
        }
        return first;
    }

    /**
     * 两个有环链表是否相交，相交返回第一个相交节点，不相交返回null
     * @param head1  第一个链表的头节点
     * @param loop1  第一个链表的入环节点
     * @param head2 第二个链表的头节点
     * @param loop2 第二个链表的入环节点
     * @return
     */
    public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
        Node cur1 = null;
        Node cur2 = null;
        if (loop1 == loop2) { //对应情况一和情况二，与判断两个无环的链表是否相交是同样的步骤，只是结束节点变成了loop1
            cur1 = head1;
            cur2 = head2;
            int n = 0;
            while (cur1 != loop1) {
                n++;
                cur1 = cur1.next;
            }
            while (cur2 != loop2) {
                n--;
                cur2 = cur2.next;
            }
            cur1 = n > 0 ? head1 : head2;
            cur2 = cur1 == head1 ? head2 : head1;
            n = Math.abs(n);
            while (n != 0) {
                n--;
                cur1 = cur1.next;
            }
            while (cur1 != cur2) {
                cur1 = cur1.next;
                cur2 = cur2.next;
            }
            return cur1;
        } else {   //对应情况三和情况四，其中情况三相交的，情况四不相交
            cur1 = loop1.next;
            while (cur1 != loop1) {
                if (cur1 == loop2) {
                    return loop1;
                }
                cur1 = cur1.next;
            }
            return null;
        }
    }

    public static void main(String[] args) {
        // 1->2->3->4->5->6->7->null
        Node head1 = new Node(1);
        head1.next = new Node(2);
        head1.next.next = new Node(3);
        head1.next.next.next = new Node(4);
        head1.next.next.next.next = new Node(5);
        head1.next.next.next.next.next = new Node(6);
        head1.next.next.next.next.next.next = new Node(7);

        // 0->9->8->6->7->null
        Node head2 = new Node(0);
        head2.next = new Node(9);
        head2.next.next = new Node(8);
        head2.next.next.next = head1.next.next.next.next.next; // 8->6
        System.out.println(getIntersectNode(head1, head2).value);

        // 1->2->3->4->5->6->7->4...
        head1 = new Node(1);
        head1.next = new Node(2);
        head1.next.next = new Node(3);
        head1.next.next.next = new Node(4);
        head1.next.next.next.next = new Node(5);
        head1.next.next.next.next.next = new Node(6);
        head1.next.next.next.next.next.next = new Node(7);
        head1.next.next.next.next.next.next.next = head1.next.next.next; // 7->4

        // 0->9->8->2...
        head2 = new Node(0);
        head2.next = new Node(9);
        head2.next.next = new Node(8);
        head2.next.next.next = head1.next; // 8->2
        System.out.println(getIntersectNode(head1, head2).value);

        // 0->9->8->6->4->5->6..
        head2 = new Node(0);
        head2.next = new Node(9);
        head2.next.next = new Node(8);
        head2.next.next.next = head1.next.next.next.next.next; // 8->6
        System.out.println(getIntersectNode(head1, head2).value);

    }

}

 

两个有环单向链表相交示例图

 

 

两个无环单向链表相交示例图