Seeking a single list penultimate node K

Seeking a single list penultimate K values

topic:

Single list to find the inverse element of K, such as to order the list: the penultimate element 2 1-> 2-> 3-> 4-> 5, the chain 4 is

• Construct a simple singly linked list

• Thinking

  • 1, the order of traversal method twice
  • 2, the speed of the pointer using the method

• Code

  • Code

1, each node comprising a single linked list pointer that points to a node data and

public class LNode {
    int data; //数据域
    LNode next;  //下一个节点的引用
}

2, the sequence is traversed twice, the first time through the entire length of n is obtained the list, the second pass of the obtained n-k + 1 th element is the reciprocal of K elements, this method requires to traverse the list twice

3, the speed of the pointer method is to use two pointers, a pointer quickly, a slow pointer, the head pointer to the first two nodes, and then quickly move the pointer K positions, this time the distance between two pointers is K, then two Meanwhile pointer movement, when the node is null pointer quickly when the pointer is the node slow penultimate node K

4, code implementation

public class _015 {

    /**
     * 快慢指针法
     * @param head 链表的头节点
     * @param k
     * @return
     */
    public static LNode FindLastK(LNode head, int k) {
        if (head == null || head.next == null)
            return head;
        LNode slow, fast;
        slow = fast = head.next;
        int i;
        for (i = 0; i < k && fast != null; ++i) {
            fast = fast.next;
        }
        if (i < k)
            return null;
        while (fast != null) {
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }

    //顺序遍历两遍法
    public static LNode findLastK1(LNode head, int k) {
        if (head == null || head.next == null)
            return head;
        LNode tmpLNode = head.next;
        int n = 0;
        while (head.next != null) {
            n++;
            head = head.next;
        }
        head.next = tmpLNode;
        int t = n - k + 1;
        while (head.next != null) {
            if (t == 0)
                return head;
            t--;
            head = head.next;
        }
        return null;
    }

    /**
     * 构造一个带有头节点的链表
     * head->1->2->3->4->5
     * @param args
     */
    public static void main(String[] args) {
        LNode head = new LNode();
        head.next = null;
        LNode tmp = null;
        LNode cur = head;

        for (int i = 1; i < 7; i++) {
            tmp = new LNode();
            tmp.data = i;
            tmp.next = null;
            cur.next = tmp;
            cur = tmp;
        }

        for (cur = head.next; cur != null; cur = cur.next) {
            System.out.print("构造的链表："+cur.data + " ");
        }
        System.out.println();
        System.out.println("快慢指针求得的数值："+FindLastK(head, 3).data);
        System.out.println("顺序遍历两遍求得的值："+findLastK1(head, 3).data);
    }

}