给定字符和单词的2D网格，任务是检查该单词在网格中是否存在。一个单词可以在任意四个方向上匹配。
这4个方向分别是：水平左右方向，垂直上下方向。
例子：
Input: grid[][] = {"axmy",
"bgdf",
"xeet",
"raks"};
Output: Yes
a x m y
b g d f
x e e t
r a k s
Input: grid[][] = {"axmy",
"brdf",
"xeet",
"rass"};
Output : No
资料来源：微软访谈
方法：下面的步骤描述了此处使用的想法：
- 检查每个单元格(如果该单元格具有第一个字符)，然后一个一个地重复出现，并尝试从该单元格的所有4个方向进行匹配。
- 将网格中的位置标记为已访问，然后在4个可能的方向上重复出现。
- 重复执行后，再次将职位标记为未访问。
- 单词中的所有字母匹配后，返回true。
下面是上述方法的实现：
C++
// C++ program to check if the word
// exists in the grid or not
#include
using namespace std;
#define r 4
#define c 5
// Function to check if a word exists in a grid
// starting from the first match in the grid
// level: index till which pattern is matched
// x, y: current position in 2D array
bool findmatch(char mat[r], string pat, int x, int y,
int nrow, int ncol, int level)
{
int l = pat.length();
// Pattern matched
if (level == l)
return true;
// Out of Boundary
if (x < 0 || y < 0 || x >= nrow || y >= ncol)
return false;
// If grid matches with a letter while
// recursion
if (mat[x][y] == pat[level]) {
// Marking this cell as visited
char temp = mat[x][y];
mat[x][y] = '#';
// finding subpattern in 4 directions
bool res = findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) |
findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) |
findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) |
findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1);
// marking this cell
// as unvisited again
mat[x][y] = temp;
return res;
}
else // Not matching then false
return false;
}
// Function to check if the word exists in the grid or not
bool checkMatch(char mat[r], string pat, int nrow, int ncol)
{
int l = pat.length();
// if total characters in matrix is
// less then pattern lenghth
if (l > nrow * ncol)
return false;
// Traverse in the grid
for (int i = 0; i < nrow; i++) {
for (int j = 0; j < ncol; j++) {
// If first letter matches, then recur and check
if (mat[i][j] == pat[0])
if (findmatch(mat, pat, i, j, nrow, ncol, 0))
return true;
}
}
return false;
}
// Driver Code
int main()
{
char grid[r] = { "axmy",
"bgdf",
"xeet",
"raks" };
// Function to check if word exists or not
if (checkMatch(grid, "geeks", r, c))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java program to check if the word
// exists in the grid or not
class GFG
{
static final int r = 4;
static final int c = 4;
// Function to check if a word exists in a grid
// starting from the first match in the grid
// level: index till which pattern is matched
// x, y: current position in 2D array
static boolean findmatch(char mat[][], String pat, int x, int y,
int nrow, int ncol, int level)
{
int l = pat.length();
// Pattern matched
if (level == l)
return true;
// Out of Boundary
if (x < 0 || y < 0 || x >= nrow || y >= ncol)
return false;
// If grid matches with a letter while
// recursion
if (mat[x][y] == pat.charAt(level))
{
// Marking this cell as visited
char temp = mat[x][y];
mat[x][y] = '#';
// finding subpattern in 4 directions
boolean res = findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) |
findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) |
findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) |
findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1);
// marking this cell
// as unvisited again
mat[x][y] = temp;
return res;
}
else // Not matching then false
return false;
}
// Function to check if the word exists in the grid or not
static boolean checkMatch(char mat[][], String pat, int nrow, int ncol)
{
int l = pat.length();
// if total characters in matrix is
// less then pattern lenghth
if (l > nrow * ncol)
return false;
// Traverse in the grid
for (int i = 0; i < nrow; i++)
{
for (int j = 0; j < ncol; j++)
{
// If first letter matches, then recur and check
if (mat[i][j] == pat.charAt(0))
if (findmatch(mat, pat, i, j, nrow, ncol, 0))
return true;
}
}
return false;
}
// Driver Code
public static void main(String[] args)
{
char grid[][] = { "axmy".toCharArray(),
"bgdf".toCharArray(),
"xeet".toCharArray(),
"raks".toCharArray() };
// Function to check if word exists or not
if (checkMatch(grid, "geeks", r, c))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program to check if the word
# exists in the grid or not
r = 4
c = 4
# Function to check if a word exists
# in a grid starting from the first
# match in the grid level: index till
# which pattern is matched x, y: current
# position in 2D array
def findmatch(mat, pat, x, y,
nrow, ncol, level) :
l = len(pat)
# Pattern matched
if (level == l) :
return True
# Out of Boundary
if (x < 0 or y < 0 or
x >= nrow or y >= ncol) :
return False
# If grid matches with a letter
# while recursion
if (mat[x][y] == pat[level]) :
# Marking this cell as visited
temp = mat[x][y]
mat[x].replace(mat[x][y], "#")
# finding subpattern in 4 directions
res = (findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) |
findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) |
findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) |
findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1))
# marking this cell as unvisited again
mat[x].replace(mat[x][y], temp)
return res
else : # Not matching then false
return False
# Function to check if the word
# exists in the grid or not
def checkMatch(mat, pat, nrow, ncol) :
l = len(pat)
# if total characters in matrix is
# less then pattern lenghth
if (l > nrow * ncol) :
return False
# Traverse in the grid
for i in range(nrow) :
for j in range(ncol) :
# If first letter matches, then
# recur and check
if (mat[i][j] == pat[0]) :
if (findmatch(mat, pat, i, j,
nrow, ncol, 0)) :
return True
return False
# Driver Code
if __name__ == "__main__" :
grid = ["axmy", "bgdf",
"xeet", "raks"]
# Function to check if word
# exists or not
if (checkMatch(grid, "geeks", r, c)) :
print("Yes")
else :
print("No")
# This code is contributed by RyugaC#
// C# program to check if the word
// exists in the grid or not
using System;
class GFG
{
static readonly int r = 4;
static readonly int c = 4;
// Function to check if a word exists in a grid
// starting from the first match in the grid
// level: index till which pattern is matched
// x, y: current position in 2D array
static bool findmatch(char [,]mat, String pat, int x, int y,
int nrow, int ncol, int level)
{
int l = pat.Length;
// Pattern matched
if (level == l)
return true;
// Out of Boundary
if (x < 0 || y < 0 || x >= nrow || y >= ncol)
return false;
// If grid matches with a letter while
// recursion
if (mat[x, y] == pat[level])
{
// Marking this cell as visited
char temp = mat[x, y];
mat[x, y] = '#';
// finding subpattern in 4 directions
bool res = findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) |
findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) |
findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) |
findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1);
// marking this cell
// as unvisited again
mat[x, y] = temp;
return res;
}
else // Not matching then false
return false;
}
// Function to check if the word exists in the grid or not
static bool checkMatch(char [,]mat, String pat, int nrow, int ncol)
{
int l = pat.Length;
// if total characters in matrix is
// less then pattern lenghth
if (l > nrow * ncol)
return false;
// Traverse in the grid
for (int i = 0; i < nrow; i++)
{
for (int j = 0; j < ncol; j++)
{
// If first letter matches, then recur and check
if (mat[i, j] == pat[0])
if (findmatch(mat, pat, i, j, nrow, ncol, 0))
return true;
}
}
return false;
}
// Driver Code
public static void Main(String[] args)
{
char [,]grid = { {'a','x','m','y'},
{'b','g','d','f'},
{'x','e','e','t'},
{'r','a','k','s'} };
// Function to check if word exists or not
if (checkMatch(grid, "geeks", r, c))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by 29AjayKumar输出：
Yes