给定一个人位于current_pos位置，并且二进制字符串路径表示该人所执行的移动，如果path [i] =’0′则该人向左移动了一步，如果path [i] =’1′则该人向左移动了向右走一步。任务是找到该人拜访的不同职位的数量。
例子：
Input: current_pos = 5, path = “011101”
Output: 4
Given moves are left, right, right, right, left and right
i.e. 5 -> 4 -> 5 -> 6 -> 7 -> 6 -> 7
The number of distinct positions are 4 (4, 5, 6 and 7).
Input: current_pos = 3, path = “110100”
Output: 3
3 -> 4 -> 5 -> 4 -> 5 -> 4 -> 3
方法：
- 声明一个数组points []以存储该人经历的所有点。
- 将此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：
- 如果当前字符为‘0’ ，那么该人向左走。因此，将当前位置减1，然后将其存储在points []中。
- 如果当前字符为‘1’ ，则表示该人向右走。因此，将当前位置增加1并将其存储在points []中。
- 计算点中不同元素的总数[] 。有关对数组中的不同元素进行计数的不同方法，请参阅对数组中的不同元素进行计数。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Utility function to return the number
// of distinct elements in an array
int countDistinct(int arr[], int len)
{
set 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
// 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 hs = new HashSet();
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));
}
} Python3
# 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 kumarC#
// 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 hs = new HashSet();
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 PHP
输出：
4