刷题篇（一）

刷题总结

第一题

最近题目刷的还是少，笔试做得惨不忍睹。
首先分享一道百度的笔试题（2023-09-27）。java编程题第三题
题目忘完了快，不过百度让用本地idea。所以记录下来了代码。
就是有关注释的最后一道编程题。当然昨天没写对。差一个条件没搞出来很伤心。

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        List<String> list=new ArrayList<>();
        while(sc.hasNextLine()){
            String line = sc.nextLine();
            list.add(line);
            if("end".equals(line)) break;
        }
        for (String s:list) {
            System.out.println(s);
        }
        int count =sc.nextInt();
        for (int i = 0; i < count; i++) {
            int command= sc.nextInt();
            int num=sc.nextInt()-1;
            if(command==1){//命令一追加注释
                String s = list.get(num);
                s.concat(sc.next());
                list.set(num,s);
            }else if(command==2){//
                String s = list.get(num);
                String[] split = s.split("//");
                list.set(num,split[0]);
            }
            else if(command==3){//把对应行变成注释
                String s = list.get(num);
                if("//".equals(s.substring(0,2))) continue;
                else {
                    String concat = "//".concat(s);
                    list.set(num,concat);
                }
            }
            else if(command==4){//取消对应行注释
                String s = list.get(num);
                if("//".equals(s.substring(0,2))){
                    s=s.substring(2);
                    list.set(num,s);
                }else{
                    continue;
                }
            }
            else if(command==5){//显示注释
                String s = list.get(num);
                String[] split = s.split("//");
                //System.out.println("5:+s->"+s+","+(split.length==1));
                if(split.length==1) {
                    System.out.println("null");
                    continue;
                }
                for (int j = 1; j < split.length; j++) {
                    if(j==split.length-2)
                        System.out.print(split[j]+"//");
                    else System.out.print(split[j]);
                }
                System.out.println();
            }

        }
    }
}

输入的是：
import java.util.*;
class Main{
public static void main(String[] args){
Node x,y=new Node(5); //‘new’ can lots of things.
//Node b=new Node();
System.out.print(x.val);
//return; //it can be ignored.
}
}
end
10
5 4
3 4
5 4
5 5
4 5
5 5
1 7 Nice!
5 7
4 7
5 7

结果

---

剑指offer

JZ3 数组中重复的数字

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param numbers int整型一维数组
     * @return int整型
     */
    public int duplicate (int[] numbers) {
        // write code here
        int length = numbers.length;
        boolean[] flag = new boolean[length];
        for (int i = 0; i < length; i++) {
            flag[i] = false;
        }
        for (int i = 0; i < length; i++) {
            if (flag[numbers[i]] == false) flag[numbers[i]] = true;
            else return numbers[i];
        }
        return -1;
    }
}

一个标记数组搞定。
JZ4 二维数组中的查找

public class Solution {
    public boolean Find(int target, int [][] array) {
        //优先判断特殊
        if(array.length == 0)  
            return false;
        int n = array.length;
        if(array[0].length == 0)  
            return false;
        int m = array[0].length;
        //从最左下角的元素开始往左或往上
        for(int i = n - 1, j = 0; i >= 0 && j < m; ){ 
            //元素较大，往上走
            if(array[i][j] > target)   
                i--;
            //元素较小，往右走
            else if(array[i][j] < target) 
                j++;
            else
                return true;
        }
        return false;
    }
}

JZ7 重建二叉树

用到了递归。
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre, int [] vin) {
        int n = pre.length;
        int m = vin.length;
        if (n == 0 || m == 0) return  null;
        TreeNode root = new TreeNode(pre[0]);
        for (int i = 0; i < vin.length; i++) {
            if (pre[0] == vin[i]) {
                root.left = reConstructBinaryTree(Arrays.copyOfRange(pre, 1, i + 1),
                                                  Arrays.copyOfRange(vin, 0, i));
                root.right = reConstructBinaryTree(Arrays.copyOfRange(pre, i + 1, pre.length),
                                                   Arrays.copyOfRange(vin, i + 1, vin.length));
                break;
            }
        }
        return root;
    }
}

JZ8 二叉树的下一个结点
中序遍历 左根右

/*
public class TreeLinkNode {
    int val;
    TreeLinkNode left = null;
    TreeLinkNode right = null;
    TreeLinkNode next = null;

    TreeLinkNode(int val) {
        this.val = val;
    }
}
*/
import java.util.*;
public class Solution {
    ArrayList<TreeLinkNode> nodes = new ArrayList<>();
    public TreeLinkNode GetNext(TreeLinkNode pNode) {
        // 获取根节点
        TreeLinkNode root = pNode;
        while (root.next != null) root = root.next;

        // 中序遍历打造nodes
        InOrder(root);

        // 进行匹配
        int n = nodes.size();
        for (int i = 0; i < n - 1; i++) {
            TreeLinkNode cur = nodes.get(i);
            if (pNode == cur) {
                return nodes.get(i + 1);
            }
        }
        return null;
    }

    // 中序遍历
    void InOrder(TreeLinkNode root) {
        if (root != null) {
            InOrder(root.left);
            nodes.add(root);
            InOrder(root.right);
        }
    }
}

JZ10 斐波那契数列
斐波那契数列
1 1 2 3 5 8
当前位置数等于前两个数之和

public class Solution {
    public int Fibonacci(int n) {
        if (n <= 2) return 1;
        int num1 = 1;
        int num2 = 1;
        int num3 = 0;
        for (int i = 3; i <= n; i++) {
            num3 = num1 + num2;
            num1 = num2;
            num2 = num3;
        }
        return num3;
    }
}