力扣labuladong——一刷day01

最新推荐文章于 2025-12-04 16:02:12 发布

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#leetcode #算法 #职场和发展 #java #数据结构

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本文介绍了四个关于链表的LeetCode题目：合并两个有序链表、分隔链表、合并K个升序链表以及删除链表的倒数第N个节点，详细展示了解决方案。

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文章目录

前言
一、21. 合并两个有序链表
二、力扣86. 分隔链表
三、力扣23. 合并 K 个升序链表
四、力扣删除链表的倒数第 N 个结点

前言

一、21. 合并两个有序链表

/**
 * 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 res = new ListNode(), pre = res;
        ListNode p1 = list1, p2 = list2;
        while(p1 != null && p2 != null){
            if(p1.val <= p2.val){
                pre.next = p1;
                p1 = p1.next;
                pre = pre.next; 
            }else{
                pre.next = p2;
                p2 = p2.next;
                pre = pre.next;
            }
        }
        while(p1 != null){
            pre.next = p1;
            p1 = p1.next;
            pre = pre.next;
        }
        while(p2 != null){
            pre.next = p2;
            p2 = p2.next;
            pre = pre.next;
        }
        return res.next;
    }
}

二、力扣86. 分隔链表

/**
 * 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 partition(ListNode head, int x) {
        ListNode dump1 = new ListNode(-1,null);
        ListNode dump2 = new ListNode(-1,null);
        ListNode p = head, p1 = dump1, p2 = dump2;
        while(p != null){
            if(p.val < x){
                p1.next = p;
                p1 = p1.next;
            }else{
                p2.next = p;
                p2 = p2.next;
            }
            ListNode temp = p.next;
            p.next = null;
            p = temp;
        }
        p1.next = dump2.next;
        return dump1.next;
    }
}

三、力扣23. 合并 K 个升序链表

/**
 * 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 mergeKLists(ListNode[] lists) {
        if(lists.length == 0){
            return null;
        }
        ListNode res = new ListNode(-1,null);
        ListNode r = res;
        PriorityQueue<ListNode> pq = new PriorityQueue<>(
            lists.length, (a,b)->(a.val - b.val)
        );
        for(ListNode node : lists){
            if(node != null){
                pq.add(node);
            }
        }
        while(!pq.isEmpty()){
            ListNode p = pq.poll();
            r.next = p;
            if(p.next != null){
                pq.add(p.next);
            }
            r = r.next;
        }
        return res.next;
    }
}

四、力扣删除链表的倒数第 N 个结点

/**
 * 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 removeNthFromEnd(ListNode head, int n) {
        ListNode p1 = head;
        for(int i = 0; i < n; i ++){
            p1 = p1.next;
        }
        if(p1 == null){
            return head.next;
        }
        ListNode p2 = head;
        while(p1.next != null){
            p1 = p1.next;
            p2 = p2.next;
        }
        p2.next = p2.next.next;
        return head;
    }
}