给定长度为N 的字符串str仅由字符‘N’、’S’、’E’或‘W’ 组成，每个字符分别代表向北、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，并按照字符串的方向行走。任务是检查该人是否在步行中穿过了他已经访问过的任何坐标。如果发现是真的，则打印“ Crossed” 。否则，打印“未交叉” 。
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来检查是否已经访问过的坐标出现。以下是步骤：
- 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。
- 在集合中插入X和Y。
- 在[0, N)范围内迭代一个循环，并根据以下条件增加 X 或 Y 坐标：
- 如果字符是‘N’ ，则将Y增加1 。
- 如果字符是‘S’ ，则将Y减1 。
- 如果字符是‘E’ ，则将X增加1 。
- 如果字符是‘W’ ，则将X减1 。
- 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。
- 在上述步骤中插入坐标时，如果遇到任何现有坐标，则打印Crossed 。
- 否则，打印Not Crossed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 divyeshrabadiya07C#
// 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)
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