检查路径序列是否两次访问任何坐标

给定长度为N的字符串str ，该字符串仅由字符‘N’，’S’，’E’或‘W’组成，每个字符分别表示分别向北，向南，向东或向西移动一个单位。一个人从2D平面上的原点(0，0)开始，然后按照字符串的方向行走。任务是检查该人是否已走过他走过的任何坐标。如果发现是正确的，则打印“划线” 。否则，请打印“ Not Crossed” 。

例子：

Input: str = “NESW”
Output: Crossed
Explanation:
The path for the man is given by:
(0, 0) -> (0, 1) -> (1, 1) -> (0, 1) -> (0, 0).
Since, the coordinate (0, 0) is visited again.
Therefore, print Crossed as he has crossed the path.

Input: str = “SESS”
Output: Not Crossed
Explanation:
The path for the man is given by:
(0, 0) -> (0, -1) -> (1, -1) -> (1, -2) -> (1, -3).
From the above path, the man never visited the same coordinate again.
Therefore, print Not Crossed.

方法：想法是为遇到的坐标维护一个Set ，以检查是否已经访问过的坐标。步骤如下：

1. 初始化两个变量X和Y并将值初始化为零。这些变量将代表x和y坐标。
2. 在集合中插入X和Y。
3. 在[0，N)范围内循环循环，并根据以下条件递增X或Y坐标：
   • 如果字符为“ N” ，则将Y递增1 。
   • 如果字符为“ S” ，则将Y减1 。
   • 如果字符为“ E” ，则将X递增1 。
   • 如果字符为“ W” ，则将X减1 。
4. 在上述步骤中，为集合中的每个字符插入更新的X和Y坐标。
5. 在上述步骤中插入坐标时，如果遇到任何当前坐标，则打印Crossed 。
6. 否则，打印“不交叉”，因为所有坐标点都是唯一的。

下面是上述方法的实现

C++

// C++ program for the above appproach
 
#include 
using namespace std;
 
// Function to check if the man crosses
// previous visited coordinate or not
bool isCrossed(string path)
{
    if (path.size() == 0)
        return false;
 
    // Stores the count of
    // crossed vertex
    bool ans = false;
 
    // Stores (x, y) coordinates
    set > set;
 
    // The coordinates for the origin
    int x = 0, y = 0;
    set.insert({ x, y });
 
    // Iterate over the string
    for (int i = 0; i < path.size(); i++) {
 
        // Condition to increment X or
        // Y co-ordinates respectively
        if (path[i] == 'N')
            set.insert({ x, y++ });
 
        if (path[i] == 'S')
            set.insert({ x, y-- });
 
        if (path[i] == 'E')
            set.insert({ x++, y });
 
        if (path[i] == 'W')
            set.insert({ x--, y });
 
        // Check if (x, y) is already
        // visited
        if (set.find({ x, y })
            != set.end()) {
 
            ans = true;
            break;
        }
    }
 
    // Print the result
    if (ans)
        cout << "Crossed";
    else
        cout << "Not Crossed";
}
 
// Driver Code
int main()
{
    // Given string
    string path = "NESW";
 
    // Function Call
    isCrossed(path);
 
    return 0;
}

Java

// Java program for
// the above appproach
import java.awt.Point;
import java.util.HashSet;
class GFG{
 
// Function to check if the man crosses
// previous visited coordinate or not
static void isCrossed(String path)
{
  if (path.length() == 0)
    return;
 
  // Stores the count of
  // crossed vertex
  boolean ans = false;
 
  // Stores (x, y) coordinates
  HashSet set =
          new HashSet();
 
  // The coordinates for the origin
  int x = 0, y = 0;
  set.add(new Point(x, y));
 
  // Iterate over the String
  for (int i = 0; i < path.length(); i++)
  {
    // Condition to increment X or
    // Y co-ordinates respectively
    if (path.charAt(i) == 'N')
      set.add(new Point(x, y++));
 
    if (path.charAt(i) == 'S')
      set.add(new Point(x, y--));
 
    if (path.charAt(i) == 'E')
      set.add(new Point(x++, y));
 
    if (path.charAt(i) == 'W')
      set.add(new Point(x--, y));
 
    // Check if (x, y) is already
    // visited
    if (set.contains(new Point(x, y)))
    {
      ans = true;
      break;
    }
  }
 
  // Print the result
  if (ans)
    System.out.print("Crossed");
  else
    System.out.print("Not Crossed");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given String
  String path = "NESW";
 
  // Function Call
  isCrossed(path);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program for the above appproach
 
# Function to check if the man crosses
# previous visited coordinate or not
def isCrossed(path):
 
    if (len(path) == 0):
        return bool(False)
 
    # Stores the count of
    # crossed vertex
    ans = bool(False)
 
    # Stores (x, y) coordinates
    Set = set()
 
    # The coordinates for the origin
    x, y = 0, 0
    Set.add((x, y))
 
    # Iterate over the string
    for i in range(len(path)):
 
        # Condition to increment X or
        # Y co-ordinates respectively
        if (path[i] == 'N'):
            Set.add((x, y))
            y = y + 1
 
        if (path[i] == 'S'):
            Set.add((x, y))
            y = y - 1
             
        if (path[i] == 'E'):
            Set.add((x, y))
            x = x + 1
 
        if (path[i] == 'W'):
            Set.add((x, y))
            x = x - 1
             
        # Check if (x, y) is already visited
        if (x, y) in Set:
            ans = bool(True)
            break
 
    # Print the result
    if (ans):
        print("Crossed")
    else:
        print("Not Crossed")
 
# Driver code
 
# Given string
path = "NESW"
 
# Function call
isCrossed(path)
 
# This code is contributed by divyeshrabadiya07

C#

// C# program for the above appproach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check if the man crosses
// previous visited coordinate or not
static void isCrossed(String path)
{
    if (path.Length == 0)
        return;
 
    // Stores the count of
    // crossed vertex
    bool ans = false;
 
    // Stores (x, y) coordinates
    HashSet> mySet =
    new HashSet>();
 
    // The coordinates for the origin
    int x = 0, y = 0;
    mySet.Add(new KeyValuePair(x, y));
 
    // Iterate over the String
    for(int i = 0; i < path.Length; i++)
    {
         
        // Condition to increment X or
        // Y co-ordinates respectively
        if (path[i] == 'N')
            mySet.Add(
                new KeyValuePair(x, y++));
 
        if (path[i] == 'S')
            mySet.Add(
                new KeyValuePair(x, y--));
 
        if (path[i] == 'E')
            mySet.Add(
                new KeyValuePair(x++, y));
 
        if (path[i] == 'W')
            mySet.Add(
                new KeyValuePair(x--, y));
 
        // Check if (x, y) is already
        // visited
        if (mySet.Contains(
                new KeyValuePair(x, y)))
        {
            ans = true;
            break;
        }
    }
 
    // Print the result
    if (ans)
        Console.Write("Crossed");
    else
        Console.Write("Not Crossed");
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String
    String path = "NESW";
 
    // Function call
    isCrossed(path);
}
}
 
// This code is contributed by akhilsaini

输出：

Crossed

时间复杂度： O(N * log N)
辅助空间： O(N)