[Java]链表练习1

1.反转一个单链表。

力扣

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode middleNode2(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while(slow != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
}

2.给定一个带有头结点 head 的非空单链表，返回链表的中间结点。如果有两个中间结点，则返回第二个中间结点。

力扣

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
}

3.输入一个链表，输出该链表中倒数第k个结点。

链表中倒数第k个结点_牛客题霸_牛客网

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        if(k <= 0 || head == null) {
            return null;
        }
        ListNode fast = head;
        ListNode slow = head;
        while(k - 1 > 0){
            if(fast.next == null){
                return null;
            }
            fast = fast.next;
            k--;
        }
        while(fast.next != null){
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}

4.将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

力扣

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode newNode = new ListNode();
        ListNode tmp = newNode;
        while(list1 != null&& list2 != null){
            if(list1.val <= list2.val){
                tmp.next = list1;
                tmp = tmp.next;
                list1 = list1.next;
            }else{
                tmp.next = list2;
                tmp = tmp.next;
                list2 = list2.next;
            }
        }
        if(list1 == null && list2 != null) {
            tmp.next = list2;
        }else{
                tmp.next = list1;
        }
        return newNode.next;
    }
}

5.编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前 。

链表分割_牛客题霸_牛客网

import java.util.*;

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Partition {
    public ListNode partition(ListNode pHead, int x) {
        // write code here
        if(pHead == null){
            return null;
        }
        ListNode cur = pHead;
        
        ListNode bs = null;
        ListNode be = null;
        ListNode as = null;
        ListNode ae = null;
        
        while(cur != null){
            if(cur.val < x){
                if(bs == null){
                    bs = cur;
                    be = cur;
                }else{
                    be.next = cur;
                    be = be.next;
                }
            }else{
                if(as == null){
                    as = cur;
                    ae = cur;
                }else{
                    ae.next = cur;
                    ae = ae.next;
                }
            }
            cur = cur.next;
        }
        if(bs == null){
            return as;
        }
        be.next = as;
        if(as != null){
            ae.next = null;
        }
        return bs;
    }
}