发现连续编号的最长序列 [英] finding longest sequence of consecutive numbers

问题描述

问题H（最长的天然接班人）：

连续两个整数是天然的接班人，如果第二个是第一个自然数（1和2是自然的继任者）的序列中的继任者。编写一个程序，读取数N其次是N个整数，然后打印连续天然接班人的最长序列的长度。

例如：

输入7 2 3 5 6 7 8 9 10输出3这是我的code，到目前为止，我不知道为什么它不能正常工作

 进口java.util.Scanner中;公共类形成机制{    公共静态无效的主要（字串[] args）{
        扫描程序扫描=新的扫描仪（System.in）;
        INT X = scan.nextInt（）;
        INT []数组=新INT [X];
        的for（int i = 0; I＆LT; array.length，我++）{
            数组[我] = scan.nextInt（）;
        }
        的System.out.println（阵列（阵列））;    }    公共静态int数组（INT []数组）{
        诠释计数= 0，温度= 0;
        的for（int i = 0; I＆LT; array.length，我++）{
            计数= 0;
            对于（INT J = I，K = I + 1; J＆LT; array.length  -  1; J ++，K ++）{
                如果（Math.abs（阵列[J]  - 阵[K]）== 1）{
                    算上++;
                }其他{
                    如果（温度＆LT; =计数）{
                        TEMP =计数;
                    }
                    打破;
                }
            }
        }
        返回温度+ 1;
    }}
 

解决方案

为什么两个循环？怎么样

 公共静态int数组（最终诠释[]数组）{
    INT lastNo = -100;
    INT maxConsecutiveNumbers = 0;
    INT currentConsecutiveNumbers = 0;    的for（int i = 0; I＆LT; array.length，我++）{
        如果（阵列[我] == lastNo + 1）{
            currentConsecutiveNumbers ++;
            maxConsecutiveNumbers = Math.max（maxConsecutiveNumbers，
                    currentConsecutiveNumbers）;
        }其他{
            currentConsecutiveNumbers = 1;
        }
        lastNo =阵列[我]
    }
    返回Math.max（maxConsecutiveNumbers，currentConsecutiveNumbers）;
}
 

Problem H (Longest Natural Successors):

Two consecutive integers are natural successors if the second is the successor of the first in the sequence of natural numbers (1 and 2 are natural successors). Write a program that reads a number N followed by N integers, and then prints the length of the longest sequence of consecutive natural successors.

Example:

Input 7 2 3 5 6 7 9 10 Output 3 this is my code so far and i have no idea why it does not work

import java.util.Scanner;

public class Conse {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int x = scan.nextInt();
        int[] array = new int[x];
        for (int i = 0; i < array.length; i++) {
            array[i] = scan.nextInt();
        }
        System.out.println(array(array));

    }

    public static int array(int[] array) {
        int count = 0, temp = 0;
        for (int i = 0; i < array.length; i++) {
            count = 0;
            for (int j = i, k = i + 1; j < array.length - 1; j++, k++) {
                if (Math.abs(array[j] - array[k]) == 1) {
                    count++;
                } else {
                    if (temp <= count) {
                        temp = count;
                    }
                    break;
                }
            }
        }
        return temp + 1;
    }

}

解决方案

Why two loops? What about

public static int array(final int[] array) {
    int lastNo = -100;
    int maxConsecutiveNumbers = 0;
    int currentConsecutiveNumbers = 0;

    for (int i = 0; i < array.length; i++) {
        if (array[i] == lastNo + 1) {
            currentConsecutiveNumbers++;
            maxConsecutiveNumbers = Math.max(maxConsecutiveNumbers,
                    currentConsecutiveNumbers);
        } else {
            currentConsecutiveNumbers = 1;
        }
        lastNo = array[i];
    }
    return Math.max(maxConsecutiveNumbers, currentConsecutiveNumbers);
}

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