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请叫我:Tom-shushu

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输了不可怕,大不了从头再来,我们还年轻---周红

链表常见的题型(java实现)

链表是面试中最常见的一种题型,因为他的每个题的代码短,短短的几行代码就可以体现出应聘者的编码能力,所以它也就成为了面试的重点。

链表常见的操作有1.打印链表的公共部分,2.删除链表的倒数第K个节点,3.翻转单向链表,4.环形约瑟夫环问题,5.判断链表是否是一个回文链表,6.两个链表生成相加链表,7.删除无序链表中重复出现的节点,8.删除指定值得节点,9.合并两个有序的单链表,10.环形链表的插入

import java.util.*;
/**********
 *@Author:Tom-shushu
 *@Description:链表问题
 *@Date:21:58 2019/10/2
 *  .--,       .--,
 * ( (  \.---./  ) )
 *  '.__/o   o\__.'
 *     {=  ^  =}
 *      >  -  <
 *     /       \
 *    //       \\
 *   //|   .   |\\
 *   "'\       /'"_.-~^`'-.
 *      \  _  /--'         `
 *    ___)( )(___
 *   (((__) (__)))    高山仰止,景行行止.虽不能至,心向往之。
 *
 **********/
public class Node {
    public int value;
    public Node head;
    public Node next;

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

    //打印链表的公共部分
    public void print(Node head1, Node head2) {
        while (head1 != null && head2 != null) {
            if (head1.value < head2.value) {
                head1 = head1.next;
            } else if (head1.value > head2.value) {
                head2 = head2.next;
            } else {
                System.out.println(head1.value);
                head1 = head1.next;
                head2 = head2.next;
            }
        }
    }

    //+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    //删除单链表的倒数第K个节点
    //版本一
    public Node remove1(Node head, int k) {
        if (head == null || k < 1) {
            return head;
        }
        Node cur = head;
        while (cur != null) {
            k--;
            cur = cur.next;
        }
        if (k == 0) {//要删除的是第一个
            head = head.next;
        }
        if (k < 0) {
            cur = head;
            while (++k != 0) {
                cur = cur.next;
            }
            cur.next = cur.next.next;
        }
        return head;
    }

    //版本二
    public Node remove2(Node head, int k) {
        if (head == null || k <= 0) {
            return null;
        }
        Node slow = head;
        Node fast = head;
        //fast 指向 k + 1
        for (int i = 1; i < k + 1; i++) {
            if (fast.next != null) {
                fast = fast.next;
            } else {
                return null;
            }
        }
        //fast指向尾部,slow指向倒数K+1,即 k 的前一个数。
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        //删除第 k 个数。
        slow = slow.next.next;
        return head;
    }

    //+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    //翻转单向链表
    //版本一
    public Node reList(Node head) {
        Node pre = null;
        Node next = null;
        while (head != null) {
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }

    //版本二
    public Node reList2(Node head) {
        if (head == null || head.next == null) {
            return head;
        }
        Node pre = head;
        Node newHead = null;
        while (pre != null) {
            Node temp = pre.next;
            pre.next = newHead;
            newHead = pre;
            pre = temp;
        }
        return newHead;
    }

    //+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    //环形约瑟夫问题
    public Node yuesefu(Node head, int m) {
        if (head == null || head.next == head || m < 1) {
            return head;
        }
        Node last = head;
        while (last.next != head) {
            last = last.next;
        }
        int count = 0;
        while (head != last) {
            if (++count == m) {
                last.next = head.next;
                count = 0;
            } else {
                last = last.next;
            }
            head = last.next;
        }
        return head;
    }

    //+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    //判断一个链表是否是回文链表
    public boolean isHuiWen(Node head) {
        Stack<Node> stack = new Stack<Node>();
        Node cur = head;
        while (cur != null) {
            stack.push(cur);
            cur = cur.next;
        }
        while (head != null) {
            if (head.value != stack.pop().value) {
                return false;
            }
            head = head.next;
        }
        return true;
    }
//+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

    //两个单链表生成相加链表
    public Node xinagjialainbiao(Node head1, Node head2) {
        Stack<Integer> stack1 = new Stack<Integer>();
        Stack<Integer> stack2 = new Stack<Integer>();
        while (head1 != null) {
            stack1.push(head1.value);
            head1 = head1.next;
        }
        while (head2 != null) {
            stack2.push(head1.value);
            head2 = head2.next;
        }
        int ca = 0;
        int n1 = 0;
        int n2 = 0;
        int n = 0;
        Node node = null;
        Node pre = null;
        while (!stack1.isEmpty() || !stack2.isEmpty()) {
            if (stack1.isEmpty()) {
                n1 = 0;
            } else {
                n1 = stack1.pop();
            }
            if (stack2.isEmpty()) {
                n2 = 0;
            } else {
                n2 = stack2.pop();
            }
            pre = node;
            node = new Node(n % 10);
            node.next = pre;
        }
        if (ca == 1) {
            pre = node;
            node = new Node(1);
            node.next = pre;
        }
        return node;
    }

    //+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    //删除无需单链表中重复出现的节点
    public void deletecf(Node head) {
        if (head == null) {
            return;
        }
        HashSet<Integer> set = new HashSet<Integer>();
        Node pre = head;
        Node cur = head.next;
        set.add(head.value);
        while (cur != null) {
            if (set.contains(cur.value)) {
                pre.next = cur.next;
            } else {
                set.add(cur.value);
                pre = cur;
            }
            cur = cur.next;
        }
    }

    //+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    //在单链表中删除指定值得节点
    public Node deletevalue(Node head, int num) {
        Stack<Node> stack = new Stack<Node>();
        while (head != null) {
            if (head.value != num) {
                stack.push(head);
            }
            head = head.next;
        }
        while (!stack.isEmpty()) {
            stack.peek().next = head;
            head = stack.pop();
        }
        return head;
    }

    //+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    //合并两个有序单链表(递归)
    public Node Merge(Node list1, Node list2) {
        if (list1 == null) {
            return list2;
        }
        if (list2 == null) {
            return list1;
        }
        if (list1.value <= list2.value) {
            list1.next = Merge(list1.next, list2);
            return list1;
        } else {
            list2.next = Merge(list1, list2.next);
            return list2;
        }
    }
    //++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++==
    //环形链表的插入
    public Node insertNum(Node head,int num){
        Node node = new Node(num);
        if(head == null){
            node.next = node;
            return node;
        }
        Node pre = head;
        Node cur = head.next;
        while (cur != head){
            if (pre.value <= num && cur.value >= num){
                break;
            }
            pre = cur;
            cur = cur.next;
        }
        pre.next = node;
        node.next = cur;
        return head.value < num ? head : node;
    }
}

 

输了不可怕,大不了从头再来,我们还年轻---周红

posted on 2019-10-04 17:16  Tom-shushu  阅读(229)  评论(0编辑  收藏  举报

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