30 day question brushing plan (II)

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Catalog

1. Find the longest string in the string

2. A number that appears more than half the times in an array

 3. Calculate candy

4. Hexadecimal conversion

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1. Find the longest string in the string

Find the longest consecutive number string in the string _ Niuke Tiba _ Cattle from 【 Niuke Tiba 】 Collect high-frequency school recruitment pen interview questions from enterprises , With official explanation , Online Baidu Alibaba Tencent Netease and other Internet famous enterprises written interview simulation test practice , Discuss the classic test questions with Niu Ren , Improve your technical ability in an all-round way https://www.nowcoder.com/practice/bd891093881d4ddf9e56e7cc8416562d?tpId=85&&tqId=29864&rp=1&ru=/activity/oj&qru=/ta/2017test/question-ranking

① Topics and examples ：

 ② Method resolution ：

This question mainly considers the final storage and temporary storage , And if you update and replace . We set two strings here , One is used to store temporary results , The other is used to store the final results . It should be noted that , When the last one is a number , The original size may change , At this time, you need to compare the size of the two strings again . Assign a large result to the string where the final result is located , At the same time, we should pay attention to the process of comparison , In essence, the comparison is ASCLL The value of the code . The detailed process is embodied in the code .

The code is as follows ：

import java.util.*;
public class Main {
     public static void main(String[] args) {
         Scanner sc=new Scanner(System.in);
         String str=sc.nextLine();
         // Define two strings , One for splicing , The other is used to show the final result 
         String tmp="";
         String res="";
         int i=0;
         for(i=0;i<str.length();i++){
             // Determine whether the current is a number  
            if(str.charAt(i)>='0'&&str.charAt(i)<='9'){
                 // The instructions are numbers , At this time, it is necessary to splice until it is not a number 
                 tmp+=str.charAt(i)+"";
             }else{// It means that this is not a number =》 It's letters or symbols 
                 // In order to welcome the next string of numbers , We need to compare the previous , Keep the longer one in the final string result 
                 if(res.length()<tmp.length()){
                     res=tmp;
                 }else{ // When placed , We should set the temporarily stored as 0, So that it can be put in later 
                 tmp="";
                 }
             }
         }
         // When the last number of the string is also a number , It is possible to change the size , So we need to judge again for the last time 
         if(i==str.length()&&res.length()<tmp.length()){
             res=tmp;
         }
         System.out.println(res);
     }
    }

2. A number that appears more than half the times in an array

A number that appears more than half the times in an array _ Niuke Tiba _ Cattle from 【 Niuke Tiba 】 Collect high-frequency school recruitment pen interview questions from enterprises , With official explanation , Online Baidu Alibaba Tencent Netease and other Internet famous enterprises written interview simulation test practice , Discuss the classic test questions with Niu Ren , Improve your technical ability in an all-round way https://www.nowcoder.com/practice/e8a1b01a2df14cb2b228b30ee6a92163?tpId=13&tqId=11181&tPage=2&rp=2&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

① Topics and examples ：

② Method resolution ：

a. Application map To solve ; utilize Map<key,value> Characteristics , We can use value And half the length of the array . It should be noted that there is a keySet() Returns the... Of the key contained in this map Set View . And then map All keys in the are stored in Set aggregate , because set With iterator , All iterations take out all keys . The code is as follows ：

import java.util.*;
public class Solution {
    public int MoreThanHalfNum_Solution(int [] array) {
        Map<Integer,Integer>map=new HashMap<>();
        int i=0;
        for(;i<array.length;i++){
            if(!map.containsKey(array[i])){
                map.put(array[i],1);
            }else{
                map.put(array[i],map.get(array[i])+1);
            }
        }
        int len=array.length/2;
         for(Integer key : map.keySet())
        {
            if(map.get(key) > len )
            {
                return key;
            }
        }
return 0;
    }
}

 b. We use the number in the middle of the array to judge . It's easy for us to think of , When an array is sorted from small to large , Then if the middle number is not what we are looking for greater than 1/2 The number of array lengths , Then this number does not exist , therefore , Let's find the value of the middle number first , utilize count Count , To compare other numbers in the array , If equal to it , be count++, Last comparison count and 1/2 Size of array , To determine whether the number exists . The code is as follows ：

import java.util.*;
public class Solution {
    public int MoreThanHalfNum_Solution(int [] array) {
        Arrays.sort(array);
        int len=array.length/2;
        int m=array[len];
        int count=0;
        for(int i=0;i<array.length;i++){
            if(array[i]==m){
                count++;
            }
        }
        if(count>=len){
            return m;
        }else{
            return 0;
        }
    }
}

c. We use modes to solve problems . We all know that mode is the number that appears most in a group of data , Then if there is no mode >1/2 Array length of , Then there can be no such number . We use mode , take result give 0 The number of subscripts , If the following number is the same , be count++; conversely count--; When count=0 when , It indicates that this number is not the final number for the time being , We re endow result New value . If in the end count>0, Then it shows that such a number may exist , Then access the array , See whether the number of times this value appears in the array is greater than 1/2 Team leader , If it is greater than, the number is returned , conversely , return 0;

import java.util.*;
public class Solution {
    public int MoreThanHalfNum_Solution(int [] array) {
      int result=array[0];
      int count=1;// here result The number in appears once ,count Write it down as 1
        for(int i=0;i<array.length;++i){
           // Guarantee result The premise that the value does not change is count！=0, therefore count It should be on the outer layer 
            if(count!=0){
                if(array[i]==result){
                    ++count;
                }// If you don't wait --
                else{
                    --count;
                }
            }else{// Here is count==0
                array[i]=result;
                count=1;           
            }
        }count=0;
            for(int i=0;i<array.length;++i){
                if(array[i]==result){
                    ++count;
                }
            }
       return (count > array.length/2) ? result : 0;
    }
}

 3. Calculate candy

Calculate candy _ Niuke Tiba _ Cattle from 【 Niuke Tiba 】 Collect high-frequency school recruitment pen interview questions from enterprises , With official explanation , Online Baidu Alibaba Tencent Netease and other Internet famous enterprises written interview simulation test practice , Discuss the classic test questions with Niu Ren , Improve your technical ability in an all-round way https://www.nowcoder.com/practice/02d8d42b197646a5bbd0a98785bb3a34?tpId=85&&tqId=29857&rp=1&ru=/activity/oj&qru=/ta/2017test/question-ranking

① Topics and examples ：

② Method resolution ：（ To put it simply, it is a problem of judging whether there is a solution to a system of ternary linear equations ）

This topic is mainly about finding rules . We can get four equations according to the known Title , If the final result is true , Then each value can only be unique , So we get the result equation through the example , It can be used as a judgment condition . The reason to use B Because B There are four times in the equation , There is a corresponding relationship between two .

  The code is as follows ：

import java.util.*;
public class Main {
 public static void main(String[]args){
     Scanner sc=new Scanner(System.in);
     int a=sc.nextInt();
     int b=sc.nextInt();
     int c=sc.nextInt();
     int d=sc.nextInt();
     int A = (a+c)/2; 
     int C = (d-b)/2; 
     int B1 = (b+d)/2; 
     int B2 = (c-a)/2; 
     if(B1 != B2) { 
         System.out.print("No"); 
     }else{
         System.out.print(A+" "+B2+" "+C);
     }       
}
}

 

4. Hexadecimal conversion

Hexadecimal conversion _ Niuke Tiba _ Cattle from 【 Niuke Tiba 】 Collect high-frequency school recruitment pen interview questions from enterprises , With official explanation , Online Baidu Alibaba Tencent Netease and other Internet famous enterprises written interview simulation test practice , Discuss the classic test questions with Niu Ren , Improve your technical ability in an all-round way https://www.nowcoder.com/practice/ac61207721a34b74b06597fe6eb67c52?tpId=85&&tqId=29862&rp=1&ru=/activity/oj&qru=/ta/2017test/question-ranking

① Topics and examples ：

 ② Method resolution ：

The following questions should be considered in this topic .

（1） The treatment of negative numbers . If the number is negative , How to handle , Here's what I'm using flg==-1 Or set it to false To process .

（2） Because the title requires , If N Greater than 9, The corresponding number rules refer to 16 Base number （ such as ,10 use A Express , wait ） So we introduce strings here , Using the relationship between remainder and subscript . For example, Yu 12, It's in 12 Position as C.

And then use it append Function to add to the end .

（3） If the front is negative , It's in append Finally, add the symbol . Last inverted string .

a. Use string to solve ： The code is as follows ：

import java.util.*;
 class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        StringBuilder sb=new StringBuilder();
        int m=sc.nextInt();
        int n= sc.nextInt();
        String ret="0123456789ABCDEF";
        // Deal with negative numbers 
        int flg=-1;
        if(m==0){
            System.out.println(0);
        }
        if(m<0){
            m=m*flg;
            flg=1;
        }
        while(m!=0){
            sb.append(ret.charAt(m%n));
            m/=n;
        }
        if(flg==1){
            sb.append("-");
        }
        sb.reverse();
        System.out.println(sb);
    }
}

b. In the above process, the stack is introduced to find , Because for the stack, it is last in first out , Therefore, there is no need to reverse the result , But we still need to add a minus sign to negative numbers , This operation should be completed before the stack . The code is as follows ：（ But this efficiency is not high , Because the stack is created again ）

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        Stack <Character>stack=new Stack<>();
        StringBuilder sb=new StringBuilder();
        int m=sc.nextInt();
        int n= sc.nextInt();
        String ret="0123456789ABCDEF";
        // Deal with negative numbers  
        int flg=-1;
        if(m==0){
            System.out.println(0);
        }
        if(m<0){
            m=m*flg;
            flg=1;
        }
        while(m!=0){
           stack.push(ret.charAt(m%n));
            m/=n;
        }
         if(flg==1){
        sb.append("-");
        }
        while(!stack.isEmpty()){
    
            sb.append(stack.pop());
        }  
        System.out.println(sb);
    }
}