java 统计数字线上访问的不同点

统计数字线上访问的不同点

原文:https://www . geesforgeks . org/count-distinct-points-按号码在线访问/

给定一个位于位置 current_pos 的人和一个二进制字符串路径，这是这个人采取的移动，如果路径[i] = '0' ，那么这个人向左移动一步，如果路径[i] = '1' ，那么这个人向右移动一步。任务是找到这个人拜访过的不同职位的数量。

示例:

输入: current_pos = 5，path = "011101" 输出: 4 给定的招式有左、右、右、右、左、右即 5->4->5->6->7->6->7 不同的位置数为 4 (4、5、6、7)。

输入: current_pos = 3，path = " 110100 " T3】输出:3 3->4->5->4->5->4->3

进场:

声明一个数组点[] 来存储这个人经历的所有点。
将该数组的第一个位置初始化为当前位置 current_pos 。
遍历字符串路径并执行以下操作:
- 如果当前人物是‘0’，则此人向左行进。所以将当前位置减 1 存储在点【】。
- 如果当前人物是‘1’，那么此人向右行进。所以将当前位置增加 1 存储在点【】。
计算点【】中不同元素的总数。参考计算数组中不同元素的数量了解计算数组中不同元素数量的不同方法。

下面是上述方法的实现:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to return the number
// of distinct elements in an array
int countDistinct(int arr[], int len)
{

    set<int> hs;

    for (int i = 0; i < len; i++) {
        // add all the elements to the HashSet
        hs.insert(arr[i]);
    }

    // Return the size of hashset as
    // it consists of all unique elements
    return hs.size();
}

// Function to return the count of
// positions the person went to
int getDistinctPoints(int current_pos, string path)
{

    // Length of path
    int len = path.length();

    // Array to store all the points traveled
    int points[len + 1];

    // The first point is the current_pos
    points[0] = current_pos;

    // For all the directions in path
    for (int i = 0; i < len; i++) {

        // Get whether the direction was left or right
        char ch = path[i];

        // If the direction is left
        if (ch == '0') {

            // Decrement the current position by 1
            current_pos--;

            // Store the current position in array
            points[i + 1] = current_pos;
        }

        // If the direction is right
        else {

            // Increment the current position by 1
            current_pos++;

            // Store the current position in array
            points[i + 1] = current_pos;
        }
    }

    return countDistinct(points, len + 1);
}

// Driver code
int main()
{
    int current_pos = 5;
    string path = "011101";

    cout << (getDistinctPoints(current_pos, path));

    return 0;
}
// contributed by Arnab Kundu

Java 语言(一种计算机语言，尤用于创建网站)

// Java implementation of the approach
import java.util.*;
class GFG {

    // Function to return the count of
    // positions the person went to
    public static int getDistinctPoints(int current_pos, String path)
    {

        // Length of path
        int len = path.length();

        // Array to store all the points traveled
        int points[] = new int[len + 1];

        // The first point is the current_pos
        points[0] = current_pos;

        // For all the directions in path
        for (int i = 0; i < len; i++) {

            // Get whether the direction was left or right
            char ch = path.charAt(i);

            // If the direction is left
            if (ch == '0') {

                // Decrement the current position by 1
                current_pos--;

                // Store the current position in array
                points[i + 1] = current_pos;
            }

            // If the direction is right
            else {

                // Increment the current position by 1
                current_pos++;

                // Store the current position in array
                points[i + 1] = current_pos;
            }
        }

        return countDistinct(points, len + 1);
    }

    // Utility function to return the number
    // of distinct elements in an array
    public static int countDistinct(int arr[], int len)
    {

        HashSet<Integer> hs = new HashSet<Integer>();

        for (int i = 0; i < len; i++) {
            // add all the elements to the HashSet
            hs.add(arr[i]);
        }

        // Return the size of hashset as
        // it consists of all unique elements
        return hs.size();
    }

    // Driver code
    public static void main(String[] args)
    {
        int current_pos = 5;
        String path = "011101";

        System.out.print(getDistinctPoints(current_pos, path));
    }
}

Python 3

# Utility function to return the number
# of distinct elements in an array
def countDistinct(arr, Len):

    hs = dict()

    for i in range(Len):

        # add all the elements to the HashSet
        hs[arr[i]] = 1

    # Return the size of hashset as
    # it consists of all unique elements
    return len(hs)

# Function to return the count of
# positions the person went to
def getDistinctPoints(current_pos, path):

    # Length of path
    Len = len(path)

    # Array to store all the points traveled
    points = [0 for i in range(Len + 1)]

    # The first pois the current_pos
    points[0] = current_pos

    # For all the directions in path
    for i in range(Len):

        # Get whether the direction
        # was left or right
        ch = path[i]

        # If the direction is left
        if (ch == '0'):

            # Decrement the current position by 1
            current_pos -= 1

            # Store the current position in array
            points[i + 1] = current_pos

        # If the direction is right
        else:

            # Increment the current position by 1
            current_pos += 1

            # Store the current position in array
            points[i + 1] = current_pos

    return countDistinct(points, Len + 1)

# Driver code
current_pos = 5
path = "011101"

print(getDistinctPoints(current_pos, path))

# This code is contributed by mohit kumar

C

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG {

    // Function to return the count of
    // positions the person went to
    public static int getDistinctPoints(int current_pos,
                                        string path)
    {

        // Length of path
        int len = path.Length;

        // Array to store all the points traveled
        int[] points = new int[len + 1];

        // The first point is the current_pos
        points[0] = current_pos;

        // For all the directions in path
        for (int i = 0; i < len; i++) {

            // Get whether the direction was left or right
            char ch = path[i];

            // If the direction is left
            if (ch == '0') {

                // Decrement the current position by 1
                current_pos--;

                // Store the current position in array
                points[i + 1] = current_pos;
            }

            // If the direction is right
            else {

                // Increment the current position by 1
                current_pos++;

                // Store the current position in array
                points[i + 1] = current_pos;
            }
        }

        return countDistinct(points, len + 1);
    }

    // Utility function to return the number
    // of distinct elements in an array
    public static int countDistinct(int[] arr, int len)
    {

        HashSet<int> hs = new HashSet<int>();

        for (int i = 0; i < len; i++) {
            // add all the elements to the HashSet
            hs.Add(arr[i]);
        }

        // Return the size of hashset as
        // it consists of all unique elements
        return hs.Count;
    }

    // Driver code
    public static void Main(string[] args)
    {
        int current_pos = 5;
        string path = "011101";

        Console.Write(getDistinctPoints(current_pos, path));
    }
}

// This code is contributed by shrikanth13

服务器端编程语言（Professional Hypertext Preprocessor 的缩写）

<?php
// PHP implementation of the approach

// Utility function to return the number
// of distinct elements in an array
function countDistinct($arr, $len)
{ 
    $hs = array();

    for ($i = 0; $i < $len; $i++)
    {
        // add all the elements to the HashSet
        array_push($hs, $arr[$i]);
    }

    $hs = array_unique($hs);

    // Return the size of hashset as
    // it consists of all unique elements
    return count($hs);
}

// Function to return the count of
// positions the person went to
function getDistinctPoints($current_pos, $path)
{

    // Length of path
    $len = strlen($path);

    // Array to store all the points traveled
    $points = array();

    // The first point is the current_pos
    $points[0] = $current_pos;

    // For all the directions in path
    for ($i = 0; $i < $len; $i++)
    {

        // Get whether the direction was left or right
        $ch = $path[$i];

        // If the direction is left
        if ($ch == '0')
        {

            // Decrement the current position by 1
            $current_pos--;

            // Store the current position in array
            $points[$i + 1] = $current_pos;
        }

        // If the direction is right
        else
        {

            // Increment the current position by 1
            $current_pos++;

            // Store the current position in array
            $points[$i + 1] = $current_pos;
        }
    }

    return countDistinct($points, $len + 1);
}

// Driver code
$current_pos = 5;
$path = "011101";

echo getDistinctPoints($current_pos, $path);

// This code is contributed by Ryuga
?>

java 描述语言

<script>

// JavaScript implementation of the approach

    // Function to return the count of
    // positions the person went to
    function getDistinctPoints(current_pos, path)
    {

        // Length of path
        let len = path.length;

        // Array to store all the points traveled
        let points =
        Array.from({length: len + 1}, (_, i) => 0);

        // The first point is the current_pos
        points[0] = current_pos;

        // For all the directions in path
        for (let i = 0; i < len; i++) {

            // Get whether the direction was left or right
            let ch = path[i];

            // If the direction is left
            if (ch == '0') {

                // Decrement the current position by 1
                current_pos--;

                // Store the current position in array
                points[i + 1] = current_pos;
            }

            // If the direction is right
            else {

                // Increment the current position by 1
                current_pos++;

                // Store the current position in array
                points[i + 1] = current_pos;
            }
        }

        return countDistinct(points, len + 1);
    }

    // Utility function to return the number
    // of distinct elements in an array
    function countDistinct(arr, len)
    {

       let hs = new Set();

        for (let i = 0; i < len; i++) {
            // add all the elements to the HashSet
            hs.add(arr[i]);
        }

        // Return the size of hashset as
        // it consists of all unique elements
        return hs.size;
    }

    // Driver code

    let current_pos = 5;
        let path = "011101";

       document.write(getDistinctPoints(current_pos, path));

</script>

Output:

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java 统计数字线上访问的不同点

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《GeeksForGeeks Java 中文教程 2022-07-14》

java 统计数字线上访问的不同点

统计数字线上访问的不同点

C++

Java 语言(一种计算机语言，尤用于创建网站)

Python 3

C

服务器端编程语言（Professional Hypertext Preprocessor 的缩写）

java 描述语言

留言

联系客服

java 查看更多书籍