链表快慢指针问题(判断两个链表是否有换、找环的入口节点、回文结构和删除倒数第n个节点)

一、判断两个链表是否有环

快的一次走两步，慢的走一步，两个相遇说明有环

public boolean isContainsRing(ListNode head) {
        if (head == null) {
            return false;
        }

        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;

            if (fast == slow) {
                return true;
            }
        }
        return false;
    }

二、找环的入口节点

快的一次走两步，慢的走一步，两个相遇，
相遇的点开始走 一个从head走 两个相遇即为入口节点

public ListNode getRingEntrance(ListNode head) {
        if (head == null) {
            return null;
        }

        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;

            if (fast == slow) {
                ListNode cur = head;
                while (fast != cur) {
                    cur = cur.next;
                    fast = fast.next;
                }
                return cur;
            }
        }

        return null;
    }

三、判断链表回文结构

快的一次走两步，慢的走一步，然后从慢的所在位置入栈，然后从head去和栈中元素比较

public boolean isReverse(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }

        Stack<Integer> stack = new Stack<>();
        while (slow != null) {
            stack.push(slow.val);
            slow = slow.next;
        }

        ListNode cur = head;
        while (!stack.isEmpty()) {
            if (cur.val != stack.pop()) {
                return false;
            }
            cur =  cur.next;
        }

        return true;
    }

四、删除链表的倒数第n个节点

先让快的走n步，如果快的为null说明要删除第一个节点
然后一起走，slow.next = slow.next.next;

public ListNode deleteKth(ListNode head, int k) {
        ListNode fast = head;
        ListNode slow = head;
        for (int i = 0; i < k; i++) {
            fast = fast.next;
        }

        if (fast == null) {
            return head.next;
        }

		//不能让最后一个节点指向最后一个节点的下一个 所以是fast.next != null
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }

        slow.next = slow.next.next;

        return head;
    }

五、逆置链表

新建一个节点，遍历原链表做头插操作

public ListNode reverseList(ListNode head) {
        ListNode newHead = null;
        ListNode cur = head;
        while (cur != null) {
            ListNode next = cur.next;

            cur.next = newHead;
            newHead = cur;

            cur = next;
        }

        return newHead;
    }