LFU缓存结构设计&&重排链表&&删除有序链表中重复的元素-I

NC94 LFU缓存结构设计

题目链接

1、解题思路

使用LinkedHashMap解决，LinkedHashMap能保证插入的顺序，能解决相同set，get次数找到最早get和set的那个key。

2、代码

import java.util.*;


public class Solution {
    private int getMinFreqKey(Map<Integer, Integer> map) {
        int count = Integer.MAX_VALUE;
        int key = 0;
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if (entry.getValue() < count) {
                count = entry.getValue();
                key = entry.getKey();
            }
        }
        return key;
    }

    public int[] LFU(int[][] operators, int k) {
        Map<Integer, Integer> map = new LinkedHashMap<>();
        Map<Integer, Integer> keyToFreq = new LinkedHashMap<>();
        List<Integer> list = new LinkedList<>();
        for (int i = 0; i < operators.length; i++) {
            int operator = operators[i][0];
            int key = operators[i][1];
            if (operator == 1) {
                int value = operators[i][2];
                if (map.containsKey(key)) {
                    map.put(key, value);
                    keyToFreq.put(key, keyToFreq.get(key) + 1);
                } else {
                    if (map.size() == k) {
                        int minKey = getMinFreqKey(keyToFreq);
                        map.remove(minKey);
                        keyToFreq.remove(minKey);
                    }
                    map.put(key, value);
                    keyToFreq.put(key, keyToFreq.getOrDefault(key, 0) + 1);
                }
            } else {
                if (map.containsKey(key)) {
                    int val = map.get(key);
                    keyToFreq.put(key, keyToFreq.get(key) + 1);
                    list.add(val);
                } else {
                    list.add(-1);
                }
            }
        }
        int[] arr = new int[list.size()];
        for (int i = 0; i < arr.length; i++) {
            arr[i] = list.get(i);
        }
        return arr;
    }
}

NC2 重排链表

题目链接

1、解题思路

找到尾部的进行截断，对后半部分采取头插逆序，然后两个链表交叉插入就行。注意，截断的时候，要把尾结点的next置空。

2、代码

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    private int getLen(ListNode head) {
        int res = 0;
        ListNode temp = head;
        while (temp != null) {
            res = res + 1;
            temp = temp.next;
        }
        return res;
    }

    public void reorderList(ListNode head) {
        int len = getLen(head);
        if(len == 0 || len == 1){
            return;
        }
        ListNode temp = head;
        ListNode dummy = new ListNode(0);
        ListNode pre = null;
        for (int i = 1; i < (len + 1) / 2; i++) {
            temp = temp.next;
        }
        pre = temp.next;
        temp.next = null;
        temp = pre;
        while (temp != null) {
            ListNode node = temp.next;
            temp.next = dummy.next;
            dummy.next = temp;
            temp = node;
        }
        ListNode head1 = dummy.next;
        temp = head;
        while (head1 != null) {
            System.out.printf("%d %d\n", temp.val, head1.val);
            ListNode node = head1.next;
            head1.next = temp.next;
            temp.next = head1;
            temp = temp.next.next;
            head1 = node;
        }
    }
}

NC25 删除有序链表中重复的元素-I

题目链接

1、解题思路

题目中没有给头结点，创造头结点，然后while循环模拟删除就可以了。

2、代码

import java.util.*;

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

public class Solution {
    /**
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    public ListNode deleteDuplicates (ListNode head) {
        // write code here
        ListNode temp = head;
        while (temp != null) {
            int val = temp.val;
            ListNode node = temp.next;
            while (node != null && node.val == val) {
                temp.next = node.next;
                node = temp.next;
            }
            temp = node;
        }
        return head;
    }
}