【Java数据结构】链表OJ提交小记

目录

1.删除链表中所有值为val的节点 

2.反转单链表 

3.返回链表的中间节点 

4.返回链表倒数第k个节点 

5.按次序合并链表

6.按值分割链表 

7.判断链表是否为回文 

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1.删除链表中所有值为val的节点 

1. 删除链表中所有值为val的节点https://leetcode.cn/problems/remove-linked-list-elements/description/

解题思路：

 解题代码：

/**
 * 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 removeElements(ListNode head, int val) {
        if(head == null) {
            return head;
        }
        ListNode prev = head;
        ListNode cur = head.next;
        //从cur开始，进行节点的比较
        while(cur != null) {
            //比较节点与给定值
            if(cur.val == val) {
                //相等
                prev.next = cur.next;   //从cur的下一个节点开始继续与val的值进行比较，如果相同，继续向后移动cur
                cur = cur.next;
            } else {
                //不相等
                prev = cur;
                cur = cur.next;
            }
        }

        if(head.val == val) {
            head = head.next;
        }
        return head;
    }
}

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2.反转单链表 

2.反转单链表https://leetcode.cn/problems/reverse-linked-list/description/

j解题思路：

 解题代码：

/**
 * 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 reverseList(ListNode head) {
        //原地反转
        if(head == null || head.next == null) {
            return head;
        }
        ListNode cur = head.next;
        head.next = null;   //反转后现在头结点指向节点的next域应为空
        while(cur != null) {
            ListNode curNext = cur.next;
            //将cur指向的节点的next域指向当前的头结点
            cur.next = head;
            //将头结点的指向修改为cur节点
            head = cur;
            //将cur指向修改为curNext的指向
            cur = curNext;
            //重复上述操作
        }
        //反转结束
        return head;
    }
}

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3.返回链表的中间节点 

3. 返回链表的中间节点https://leetcode.cn/problems/middle-of-the-linked-list/description/

 解题思路：

 解题代码：

/**
 * 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) {
        if(head == null || head.next == null) {
            return head;
        }
        //定义快慢引用，初始化为引用head节点
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && fast.next != null) {
            //fast向后移动两次
            fast = fast.next.next;
            //slow向后移动一次
            slow = slow.next;
        }
        return slow;
    }
}

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4.返回链表倒数第k个节点 

4. 返回链表倒数第K个节点https://www.nowcoder.com/practice/529d3ae5a407492994ad2a246518148a?tpId=13&&tqId=11167&rp=2&ru=/activity/oj&qru=/ta/coding-interviews/question-ranking

解题思路：

解题代码：

    /*
    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(head == null || k <= 0) {
                return null;
            }
            ListNode first = head;
            ListNode second = head;
            for(int i=0;i

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5.按次序合并链表

5. 按照升序合并两个升序链表（不能创建新的节点）https://leetcode.cn/problems/merge-two-sorted-lists/description/

解题思路：

 解题代码

/**
 * 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) {
        if(list1 == null) {
            return list2;
        }
        if(list2 == null) {
            return list1;
        }
        //定义一个新的头节点。组装链表
        ListNode newHead = new ListNode();
        ListNode cur = newHead;
        while(list1 != null && list2 != null) {
            if(list1.val > list2.val) {
                cur.next = list2;
                list2 = list2.next;
            } else {
                cur.next = list1;
                list1 = list1.next;                
            }
            cur = cur.next;
        }
        if(list1 != null) {
            cur.next = list1;
        } else if(list2 != null) {
            cur.next = list2;
        }
        return newHead.next;
    }
}

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6.按值分割链表 

6. 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https://www.nowcoder.com/practice/0e27e0b064de4eacac178676ef9c9d70?tpId=8&&tqId=11004&rp=2&ru=/activity/oj&qru=/ta/cracking-the-coding-interview/question-ranking

解题思路：

解题代码：

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) {
        if (pHead == null || pHead.next == null) {
            return pHead;
        }
        ListNode smallHead = new ListNode(-1);
        ListNode bigHead = new ListNode(-1);
        ListNode curSamll = smallHead;
        ListNode curBig = bigHead;
        
        //进行值的比较，节点的连接
        while(pHead != null) {
            if(pHead.val < x) {
                curSamll.next = pHead;
                curSamll = pHead;
            } else {
                curBig.next = pHead;
                curBig = pHead;
            }
            pHead = pHead.next;
        }
        //连接结束后，应当将bigHead这条链表的尾节点的next域置为空。防止出现死循环

        curBig.next = null;
        //连接smallHead和bigHead这两条链表
        curSamll.next = bigHead.next;
        return smallHead.next;
    }
}

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7.判断链表是否为回文 

7.判断链表是否为回文链表http:// https://www.nowcoder.com/practice/d281619e4b3e4a60a2cc66ea32855bfa?tpId=49&&tqId=29370&rp=1&ru=/activity/oj&qru=/ta/2016test/question-ranking

解题思路：

解题代码：

import java.util.*;

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

    ListNode(int val) {2
        this.val = val;
    }
}*/
public class PalindromeList {
    public boolean chkPalindrome(ListNode head) {
        if (head == null || head.next == null) {
            return true;
        }

        //寻找链表的中间节点
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        //slow即为中间节点
        //从中间节点开始反转链表
        ListNode cur = slow.next;
        while(cur != null) {
            ListNode curNext = cur.next;
            cur.next = slow;
            slow = cur;
            cur = curNext;
        }
        //反转链表结束

        //从首个节点开始与从反转后的中间节点开始进行比较，判断是否为回文
        while(head.next != slow.next) {
            if(head.next == slow) {
                return true;
            }
            if(head.val != slow.val) {
                return false;
            }
            head = head.next;
            slow = slow.next;
        }
        return true;
    }
}