找出数组中的最大连续数字,输出数字以及连续数字的数量。

huangapple go评论121阅读模式
英文:

Find largest consecutive numbers in array and output numbers and how many there is

问题

我的代码如下打印出连续数字的数量。然而,我希望能够打印出数量以及这些数字是什么。

例如:
array = [1, 4, 9, 5, 2, 6]

这将输出:

连续数字的数量为:3

连续数字:[4 5 6]

public static int consecutive(int[] a)
{
    HashSet<Integer> values = new HashSet<Integer>();
    for (int i : a)
    {
        values.add(i);
    }
    int max = 0;
    for (int i : values) {
        if (values.contains(i - 1))
        {
            continue;
        }
        int length = 0;
        while (values.contains(i++))
        {
            length++;
        }
        max = Math.max(max, length);
    }

    return max;
}
英文:

My code below prints out how many consecutive numbers there is. However, I'm looking to print out how many there is as well as what these numbers are.

e.g.
array = [1, 4, 9, 5, 2, 6]

This would output:

Number of consecutive numbers is: 3

Consecutive numbers: [4 5 6]

public static int consecutive(int[] a)
    {
        HashSet&lt;Integer&gt; values = new HashSet&lt;Integer&gt;();
        for (int i :a)
        {
            values.add(i);
        }
        int max = 0;
        for (int i : values) {
            if (values.contains(i - 1))
            {
                continue;
            }
            int length = 0;
            while (values.contains(i++))
            {
                length++;
            }
            max = Math.max(max, length);
        }

        return max;
    }

答案1

得分: 1

按照以下方式进行操作:

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.Set;
import java.util.TreeSet;

public class Main {
    public static void main(String[] args) {
        // Tests
        List<Integer> longestConsecutive;

        longestConsecutive = longestConsecutiveList(new int[] { 1, 4, 9, 5, 2, 6 });
        System.out.println("最长连续整数列表:" + longestConsecutive + ",数量:" + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 2, 10, 4, 1, 5, 7, 3 });
        System.out.println("最长连续整数列表:" + longestConsecutive + ",数量:" + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 5, 9, 7, 10, 11, 15, 12, 4, 6 });
        System.out.println("最长连续整数列表:" + longestConsecutive + ",数量:" + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 10, 24, 20, 30, 23, 40, 25, 10, 2, 11, 3, 12 });
        System.out.println("最长连续整数列表:" + longestConsecutive + ",数量:" + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 9, 7, 3, 8, 1 });
        System.out.println("最长连续整数列表:" + longestConsecutive + ",数量:" + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 9 });
        System.out.println("最长连续整数列表:" + longestConsecutive + ",数量:" + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 1, 2 });
        System.out.println("最长连续整数列表:" + longestConsecutive + ",数量:" + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 1, 2, 3 });
        System.out.println("最长连续整数列表:" + longestConsecutive + ",数量:" + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(null);
        System.out.println("最长连续整数列表:" + longestConsecutive + ",数量:" + longestConsecutive.size());

    }

    public static List<Integer> longestConsecutiveList(int[] a) {
        if (a == null) {
            return new ArrayList<Integer>();
        }
        Set<Integer> values = new TreeSet<Integer>();
        List<Integer> list = new ArrayList<Integer>();
        List<Integer> tempList = new ArrayList<Integer>();
        int value = 0, temp = 0;

        // 将数组的元素添加到有序集合中
        for (int i : a) {
            values.add(i);
        }

        // 创建迭代器以遍历有序集合
        Iterator<Integer> itr = values.iterator();

        // 从有序集合中获取第一个元素,将其赋值给value,并将其添加到tempList中。因为tempList只有一个元素
        if (itr.hasNext()) {
            value = itr.next();
            tempList.add(value);
        }

        // 遍历有序集合的剩余元素(从第二个元素开始)
        while (itr.hasNext()) {
            // 从有序集合中获取下一个元素,将其赋值给temp
            temp = itr.next();

            // 如果temp - value = 1,则将temp添加到tempList中
            if (temp - value == 1) {
                tempList.add(temp);
            } else if (tempList.size() >= list.size()) {
                list = tempList;
                tempList = new ArrayList<Integer>();
                tempList.add(temp);
            } else {
                tempList = new ArrayList<Integer>();
            }
            value = temp;
        }
        return list.size() > tempList.size() ? list : tempList;
    }
}

输出:

最长连续整数列表:[4, 5, 6],数量:3
最长连续整数列表:[1, 2, 3, 4, 5],数量:5
最长连续整数列表:[9, 10, 11, 12],数量:4
最长连续整数列表:[10, 11, 12],数量:3
最长连续整数列表:[7, 8, 9],数量:3
最长连续整数列表:[9],数量:1
最长连续整数列表:[1, 2],数量:2
最长连续整数列表:[1, 2, 3],数量:3
最长连续整数列表:[],数量:0

代码中有详细的注释,如有疑问或问题,请随时提问。

英文:

Do it as follows:

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.Set;
import java.util.TreeSet;
public class Main {
public static void main(String[] args) {
// Tests
List&lt;Integer&gt; longestConsecutive;
longestConsecutive = longestConsecutiveList(new int[] { 1, 4, 9, 5, 2, 6 });
System.out.println(
&quot;Logest list of consecutive integers: &quot; + longestConsecutive + &quot;, Count: &quot; + longestConsecutive.size());
longestConsecutive = longestConsecutiveList(new int[] { 2, 10, 4, 1, 5, 7, 3 });
System.out.println(
&quot;Logest list of consecutive integers: &quot; + longestConsecutive + &quot;, Count: &quot; + longestConsecutive.size());
longestConsecutive = longestConsecutiveList(new int[] { 5, 9, 7, 10, 11, 15, 12, 4, 6 });
System.out.println(
&quot;Logest list of consecutive integers: &quot; + longestConsecutive + &quot;, Count: &quot; + longestConsecutive.size());
longestConsecutive = longestConsecutiveList(new int[] { 10, 24, 20, 30, 23, 40, 25, 10, 2, 11, 3, 12 });
System.out.println(
&quot;Logest list of consecutive integers: &quot; + longestConsecutive + &quot;, Count: &quot; + longestConsecutive.size());
longestConsecutive = longestConsecutiveList(new int[] { 9, 7, 3, 8, 1 });
System.out.println(
&quot;Logest list of consecutive integers: &quot; + longestConsecutive + &quot;, Count: &quot; + longestConsecutive.size());
longestConsecutive = longestConsecutiveList(new int[] { 9 });
System.out.println(
&quot;Logest list of consecutive integers: &quot; + longestConsecutive + &quot;, Count: &quot; + longestConsecutive.size());
longestConsecutive = longestConsecutiveList(new int[] { 1, 2 });
System.out.println(
&quot;Logest list of consecutive integers: &quot; + longestConsecutive + &quot;, Count: &quot; + longestConsecutive.size());
longestConsecutive = longestConsecutiveList(new int[] { 1, 2, 3 });
System.out.println(
&quot;Logest list of consecutive integers: &quot; + longestConsecutive + &quot;, Count: &quot; + longestConsecutive.size());
longestConsecutive = longestConsecutiveList(null);
System.out.println(
&quot;Logest list of consecutive integers: &quot; + longestConsecutive + &quot;, Count: &quot; + longestConsecutive.size());
}
public static List&lt;Integer&gt; longestConsecutiveList(int[] a) {
if (a == null) {
return new ArrayList&lt;Integer&gt;();
}
Set&lt;Integer&gt; values = new TreeSet&lt;Integer&gt;();
List&lt;Integer&gt; list = new ArrayList&lt;Integer&gt;();
List&lt;Integer&gt; tempList = new ArrayList&lt;Integer&gt;();
int value = 0, temp = 0;
// Add the elements of the array to the sorted set
for (int i : a) {
values.add(i);
}
// Create an iterator to navigate the sorted set
Iterator&lt;Integer&gt; itr = values.iterator();
// Get the first element from the sorted set, assign it to value and add it to
// tempList. Since tempList has one element
if (itr.hasNext()) {
value = itr.next();
tempList.add(value);
}
// Navigate the rest (2nd element onwards) of the sorted set
while (itr.hasNext()) {
// Get the next element from the sorted set and assign it to temp
temp = itr.next();
// If temp - value = 1, add temp to tempList
if (temp - value == 1) {
tempList.add(temp);
} else if (tempList.size() &gt;= list.size()) {
list = tempList;
tempList = new ArrayList&lt;Integer&gt;();
tempList.add(temp);
} else {
tempList = new ArrayList&lt;Integer&gt;();
}
value = temp;
}
return list.size() &gt; tempList.size() ? list : tempList;
}
}

Output:

Logest list of consecutive integers: [4, 5, 6], Count: 3
Logest list of consecutive integers: [1, 2, 3, 4, 5], Count: 5
Logest list of consecutive integers: [9, 10, 11, 12], Count: 4
Logest list of consecutive integers: [10, 11, 12], Count: 3
Logest list of consecutive integers: [7, 8, 9], Count: 3
Logest list of consecutive integers: [9], Count: 1
Logest list of consecutive integers: [1, 2], Count: 2
Logest list of consecutive integers: [1, 2, 3], Count: 3
Logest list of consecutive integers: [], Count: 0

I have put enough comments in the code for easy understanding. Feel free to comment in case of any doubt/issue.

答案2

得分: 0

private static int consecutive(int[] a) {
    Set<Integer> values;

    // 用于存储连续数字的列表
    List<Integer> nums = new ArrayList<>();

    values = Arrays.stream(a).boxed().collect(Collectors.toSet());

    int max = 0;
    for (int i : values) {
        if (values.contains(i - 1)) {
            continue;
        }

        // 用于存储每个序列的内部列表
        List<Integer> temp = new ArrayList<>();

        // 将 i++ 移到循环内部,因为需要将值存储起来
        while (values.contains(i)) {
            temp.add(i);
            i++;
        }

        // 如果内部列表较大,则进行替换
        if (nums.size() <= temp.size()) {
            nums = temp;
        }

        max = Math.max(max, temp.size());
    }

    System.out.println(nums);
    return max;
}
英文:
private static int consecutive(int[] a) {
Set&lt;Integer&gt; values;
// to store the consecutive numbers
List&lt;Integer&gt; nums = new ArrayList&lt;&gt;();
values = Arrays.stream(a).boxed().collect(Collectors.toSet());
int max = 0;
for (int i : values) {
if (values.contains(i - 1)) {
continue;
}
// inner list for each sequemce
List&lt;Integer&gt; temp = new ArrayList&lt;&gt;();
// moved i++ inside the loop because the value is required to store
while (values.contains(i)) {
temp.add(i);
i++;
}
// if the inner list is larger, replace
if (nums.size() &lt;= temp.size()) {
nums = temp;
}
max = Math.max(max, temp.size());
}
System.out.println(nums);
return max;
}

huangapple
  • 本文由 发表于 2020年4月4日 02:39:37
  • 转载请务必保留本文链接:https://go.coder-hub.com/61018423.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定