207研究所编程题目

207研究所编程题目

1. 数组旋转（按列打印）

package org.agic.com207_2;
public class Solution_1_matrixRotate {
    public static void main(String[] args) {
        int[][] arr = {
                {1, 2, 3},
                {4, 5, 6}
        };
        rotate(arr, 2, 3);
        
    }
    
    public static int[] rotate(int[][] arr, int rows, int cols) {
        for (int i = 0; i < cols; i++) {
            for (int j = 0; j < rows; j++) {
                System.out.print(arr[j][i] + " ");
            }
            System.out.println();
        }
        return null;
    }
}

2. 16进制的字符串转换为10进制字符串（不得调用本身的库，但是可以调用字符串处理工具）

package org.agic.com207_2;

import java.util.HashMap;
import java.util.Map;
public class Solution_2_hex2Dec {
    public static void main(String[] args) {
        System.out.println(hex2Dec("0x21")); // 33
        System.out.println(hex2Dec("10")); // 16
        System.out.println(hex2Dec("x1")); // -1
    }
    
    private static Map<Character, Integer> map = new HashMap() {
        {
            put('0', 0);
            put('1', 1);
            put('2', 2);
            put('3', 3);
            put('4', 4);
            put('5', 5);
            put('6', 6);
            put('7', 7);
            put('8', 8);
            put('9', 9);
            put('a', 10);
            put('A', 10);
            put('b', 11);
            put('B', 11);
            put('c', 12);
            put('C', 12);
            put('d', 13);
            put('D', 13);
            put('e', 14);
            put('E', 14);
            put('f', 15);
            put('F', 15);
        }
    };
    
    public static String hex2Dec(String hex) {
        try {
            String hex2 = hex.toUpperCase().replaceAll("0X", "");
            final int baseVal = 16;
            int length = hex2.length();
            int dec = 0;
            for (int i = length - 1; i >= 0; i--) {
                int power = length - i - 1;
                int x = map.get(hex2.charAt(i)).intValue() * (int) Math.pow(baseVal, power);
                dec += x;
            }
            return String.valueOf(dec);
        } catch (Exception e) {
            return "error input:" + hex;
        }
    }
}

3. 找出重复的数字，按照大小排列（方法很多）

package org.agic.com207_2;

import java.util.*;
public class Solution_3_repeatNumbers {
    public static void main(String[] args) {
        System.out.println(deleteRepeatNumbers(new int[]{1, 4, 6, 7, 2, 3, 4, 2, 3, 4}));
    }
    
    
    public static List<Integer> deleteRepeatNumbers(int[] arr) {
        Map<String, Integer> map = new HashMap<>();
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < arr.length; i++) {
            String key = String.valueOf(arr[i]);
            if (map.get(key) == null) {
                map.put(key, 1);
            } else {
                map.put(key, map.get(key).intValue() + 1);
            }
        }
        for (String key : map.keySet()) {
            if (map.get(key).intValue() > 1) {
                list.add(Integer.parseInt(key));
            }
        }
        Collections.sort(list);
        return list;
    }
}

4. 给定日期，获取该日期是当年的第几天

package org.agic.com207_2;
public class Solution_4_dayOfYear {
    public static void main(String[] args) {
        System.out.println(calcDayOfYear(2020, 3, 28));
    }
    
    private static boolean isLeap(int year) {
        return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
    }
    
    public static int calcDayOfYear(int yyyy, int MM, int dd) {
        int[] months = {0, 31, isLeap(yyyy) ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        int days = dd;
        for (int i = 1; i < MM; i++) {
            days += months[i];
        }
        return days;
    }
}

5. 长度为length的绳子拆成n段，n1+n2+n3.+… =length； n1 等均为整数，求n1n2n3… = max，求max，并且输出 n1,n2,n3…

package org.agic.com207_2;
public class Solution_5_maxMutil {
    public static void main(String[] args) {
        System.out.println(maxMutil(8, 3));
    }
    
    public static int maxMutil(int x, int n) {
        // 2    minLength: n - yLength
        // 3 3  yLength:2
        int minLength = x / n, yLength = x % n;
        return (int) (Math.pow(minLength, n - yLength)) * (int) (Math.pow(minLength + 1, yLength));
    }
}

题目难度 一星

本文地址：https://blog.csdn.net/xiaoweitianshi001/article/details/109612236