给定一个由字符的矩阵板和一个字符串Word ，任务是检查在仅由水平和垂直相邻的字符序列构成的板上是否存在Word 。每个字符只能使用一次。
例子：
Input:
board = { {‘A’, ‘B’, ‘C’, ‘E’}, {‘S’, ‘F’, ‘C’, ‘S’}, {‘A’, ‘D’, ‘E’, ‘E’} }
Word = “SEE”
Output: True
Explanation: “SEE” can be formed using characters at (1, 3)[S], (2, 3)[E] and (2, 2)[E].
Input:
board = { {‘A’, ‘B’, ‘C’, ‘E’}, {‘S’, ‘F’, ‘C’, ‘S’}, {‘A’, ‘D’, ‘E’, ‘E’} }
Word = “ABCB”
Output: False
Explanation: “ABCB” can not be formed by using adjacent characters without repetition.
方法：解决此问题的方法是遍历矩阵中的所有字符并查找单词第一个字符的出现。只要找到，就递归地继续检查其相邻的水平和垂直单元格是否存在下一个字符。重复此过程，直到找到所有字符。找到所有字符匹配表示Word的任何实例。如果没有发生这种情况，则找不到Word 。
下面是上述逻辑的实现：
C++
// C++ Program to check if a given
// word can be formed from the
// adjacent characters in a matrix
// of characters
#include
using namespace std;
// Function to check if the word exists
bool checkWord(vector >& board,
string& word, int index,
int row, int col)
{
// If index exceeds board range
if (row < 0 || col < 0
|| row >= board.size()
|| col >= board[0].size())
return false;
// If the current cell does not
// contain the required character
if (board[row][col] != word[index])
return false;
// If the cell contains the required
// character and is the last character
// of the word required to be matched
else if (index == word.size() - 1)
// Return true as word is found
return true;
char temp = board[row][col];
// Mark cell visited
board[row][col] = '*';
// Check Adjacent cells
// for the next character
if (checkWord(board, word,
index + 1, row + 1, col)
|| checkWord(board, word,
index + 1, row - 1, col)
|| checkWord(board, word,
index + 1, row, col + 1)
|| checkWord(board, word,
index + 1, row, col - 1)) {
board[row][col] = temp;
return true;
}
// Restore cell value
board[row][col] = temp;
return false;
}
// Driver Code
int main()
{
vector > board
= { { 'A', 'B', 'C', 'E' },
{ 'S', 'F', 'C', 'S' },
{ 'A', 'D', 'E', 'E' } };
string word = "CFDASABCESEE";
for (int i = 0; i < board.size(); i++) {
for (int j = 0; j < board[0].size(); j++) {
if (board[i][j] == word[0]
&& checkWord(
board, word,
0, i, j)) {
cout << "True" << '\n';
return 0;
}
}
}
cout << "False" << '\n';
return 0;
} Java
// Java Program to check if a given
// word can be formed from the
// adjacent characters in a matrix
// of characters
import java.util.*;
class GFG{
// Function to check if the word exists
static boolean checkWord(char [][]board,
String word, int index,
int row, int col)
{
// If index exceeds board range
if (row < 0 || col < 0 ||
row >= board.length ||
col >= board[0].length)
return false;
// If the current cell does not
// contain the required character
if (board[row][col] != word.charAt(index))
return false;
// If the cell contains the required
// character and is the last character
// of the word required to be matched
else if (index == word.length() - 1)
// Return true as word is found
return true;
char temp = board[row][col];
// Mark cell visited
board[row][col] = '*';
// Check Adjacent cells
// for the next character
if (checkWord(board, word,
index + 1, row + 1, col) ||
checkWord(board, word,
index + 1, row - 1, col) ||
checkWord(board, word,
index + 1, row, col + 1) ||
checkWord(board, word,
index + 1, row, col - 1))
{
board[row][col] = temp;
return true;
}
// Restore cell value
board[row][col] = temp;
return false;
}
// Driver Code
public static void main(String[] args)
{
char[][] board = { { 'A', 'B', 'C', 'E' },
{ 'S', 'F', 'C', 'S' },
{ 'A', 'D', 'E', 'E' } };
String word = "CFDASABCESEE";
for (int i = 0; i < board.length; i++)
{
for (int j = 0; j < board[0].length; j++)
{
if (board[i][j] == word.charAt(0) &&
checkWord(board, word, 0, i, j))
{
System.out.println("True");
return;
}
}
}
System.out.println("False");
}
}
// This code is contributed by Rajput-JiPython3
# Python 3 Program to check if a given
# word can be formed from the
# adjacent characters in a matrix
# of characters
# Function to check if
# the word exists
def checkWord(board, word,
index, row, col):
# If index exceeds board range
if (row < 0 or col < 0 or
row >= len(board) or
col >= len(board[0])):
return False
# If the current cell does not
# contain the required character
if (board[row][col] != word[index]):
return False
# If the cell contains the required
#character and is the last character
# of the word required to be matched
elif (index == len(word) - 1):
# Return true as word is found
return True
temp = board[row][col]
# Mark cell visited
board[row][col] = '*'
# Check Adjacent cells
# for the next character
if (checkWord(board, word,
index + 1,
row + 1, col) or
checkWord(board, word,
index + 1,
row - 1, col) or
checkWord(board, word,
index + 1,
row, col + 1) or
checkWord(board, word,
index + 1,
row, col - 1)):
board[row][col] = temp
return True
# Restore cell value
board[row][col] = temp
return False
# Driver Code
if __name__ == "__main__":
board = [['A', 'B', 'C', 'E'],
['S', 'F', 'C', 'S'],
['A', 'D', 'E', 'E']]
word = "CFDASABCESEE"
f = 0
for i in range (len(board)):
for j in range (len(board[0])):
if (board[i][j] == word[0] and
checkWord(board, word,
0, i, j)):
print ("True" )
f = 1
break
if f == 1:
break
if f == 0:
print ("False")
# This code is contributed by ChitranayalC#
// C# program to check if a given word
// can be formed from the adjacent
// characters in a matrix of characters
using System;
class GFG{
// Function to check if the word exists
static bool checkWord(char [,]board, String word,
int index, int row, int col)
{
// If index exceeds board range
if (row < 0 || col < 0 ||
row >= board.GetLength(0) ||
col >= board.GetLength(1))
return false;
// If the current cell does not
// contain the required character
if (board[row, col] != word[index])
return false;
// If the cell contains the required
// character and is the last character
// of the word required to be matched
else if (index == word.Length - 1)
// Return true as word is found
return true;
char temp = board[row, col];
// Mark cell visited
board[row, col] = '*';
// Check adjacent cells
// for the next character
if (checkWord(board, word,
index + 1, row + 1, col) ||
checkWord(board, word,
index + 1, row - 1, col) ||
checkWord(board, word,
index + 1, row, col + 1) ||
checkWord(board, word,
index + 1, row, col - 1))
{
board[row, col] = temp;
return true;
}
// Restore cell value
board[row, col] = temp;
return false;
}
// Driver Code
public static void Main(String[] args)
{
char[,] board = { { 'A', 'B', 'C', 'E' },
{ 'S', 'F', 'C', 'S' },
{ 'A', 'D', 'E', 'E' } };
String word = "CFDASABCESEE";
for(int i = 0; i < board.GetLength(0); i++)
{
for(int j = 0; j < board.GetLength(1); j++)
{
if (board[i, j] == word[0] &&
checkWord(board, word, 0, i, j))
{
Console.WriteLine("True");
return;
}
}
}
Console.WriteLine("False");
}
}
// This code is contributed by Rajput-Ji输出：
True