package data_structure;/** * author: cdfuhuichao * date: 2019/3/5 18:11 * description: 使用单链表实现 判断水仙花字符串 */public class SXHString {    public static void main(String[] args) {        System.out.println(check("上海自来水来自海上"));        System.out.println(check("上海自来来自海上"));        System.out.println(check("上"));    }    // 打印链表    public static String printNode(Node n) {        StringBuilder sb = new StringBuilder();        for (; ; ) {            sb.append(n.item.toString());            if (!n.hasNext()) {                break;            }            n = n.next;        }        System.out.println(sb.toString());        return sb.toString();    }    private static class Node
    
      {        E item;        Node
     
       next;        Node(E element) {            this.item = element;        }        public boolean hasNext() {            return next != null;        }    }    public static Node init(String s) {        // 初始化链表        Node old = new Node(s.charAt(0));        Node current = old;        for (int i = 1; i < s.length(); i++) {            current = current.next = new Node(s.charAt(i));        }        return old;    }    public static boolean check(String s) {        if (s.length() < 2) {            return true;        }        // 初始化链表        Node old = init(s);        // 快慢指针        int m = 0;        Node current = old;        Node prev = null;        Node next = null;        Node newone1 = null;        Node newone2 = null;        for (; ; ) {            boolean hasNext = current.hasNext();            next = current.next;            // 判断奇偶，快指针到终点前，将链表逆序            if ((s.length() % 2 == 0 && m < s.length()) || (s.length() % 2 != 0 && m < s.length() - 1)) {                if (prev == null) {                    current.next = null;                } else {                    current.next = prev;                }            } else {                // 快指针到终点后，将前半截链表和后半截链表拆分                if (m == s.length() - 1) {                    current = next;                }                newone1 = prev;                newone2 = current;                break;            }            m += 2;            prev = current;            current = next;            if (!hasNext) {                break;            }        }        if (newone1 == null || newone2 == null) {            return false;        }        String ss1 = printNode(newone1);        String ss2 = printNode(newone2);        return ss1.equals(ss2);    }}
     
    

欢迎指出不足和错误，谢谢。