给定大小为9×9的二维数组board [] [] ，它表示数独难题的解决方案，任务是检查所求解的数独难题的给定表示形式是否有效。
例子：
Input:
Output: Valid
Input:
Output: Not Valid
方法：通过检查以下条件可以解决该问题：
- 检查板[] []数组存储的每一行从[1,9]或不在该范围只有唯一的值。
- 检查是否从该范围的板[] []数组存储唯一值的每列[1,9]或没有。
- 检查板[] []数组存储的唯一值的所有可能的3×3子矩阵从范围[1,9]或没有。
请按照以下步骤解决问题：
- 遍历给定的矩阵board [] [] 。
- 检查是否满足上述条件。
- 如果不满足以上任何条件，则打印“无效”。
- 否则，打印“有效”。
下面是上述方法的实现：
C++
board[][] = {{7, 9, 2, 1, 5, 4, 3, 8, 6},
{6, 4, 3, 8, 2, 7, 1, 5, 9},
{8, 5, 1, 3, 9, 6, 7, 2, 4},
{2, 6, 5, 9, 7, 3, 8, 4, 1},
{4, 8, 9, 5, 6, 1, 2, 7, 3},
{3, 1, 7, 4, 8, 2, 9, 6, 5},
{1, 3, 6, 7, 4, 8, 5, 9, 2},
{9, 7, 4, 2, 1, 5, 6, 3, 8},
{5, 2, 8, 6, 3, 9, 4, 1, 7}}
Java
board[][] = {{5, 5, 5, 5, 5, 5, 5, 5, 5},
{5, 5, 5, 5, 5, 5, 5, 5, 5},
{5, 5, 5, 5, 5, 5, 5, 5, 5},
{5, 5, 5, 5, 5, 5, 5, 5, 5},
{5, 5, 5, 5, 5, 5, 5, 5, 5},
{5, 5, 5, 5, 5, 5, 5, 5, 5},
{5, 5, 5, 5, 5, 5, 5, 5, 5},
{5, 5, 5, 5, 5, 5, 5, 5, 5},
{5, 5, 5, 5, 5, 5, 5, 5, 5}}
Python3
// C++ program to implement
// the above approach
#include
using namespace std;
#define N 9
// Function to check if all elements
// of the board[][] array store
// value in the range[1, 9]
bool isinRange(int board[][N])
{
// Traverse board[][] array
for (int i = 0; i < N;
i++) {
for (int j = 0; j < N;
j++) {
// Check if board[i][j]
// lies in the range
if (board[i][j] <= 0 || board[i][j] > 9) {
return false;
}
}
}
return true;
}
// Function to check if the solution
// of sudoku puzzle is valid or not
bool isValidSudoku(int board[][N])
{
// Check if all elements of board[][]
// stores value in the range[1, 9]
if (isinRange(board)
== false) {
return false;
}
// Stores unique value
// from 1 to N
bool unique[N + 1];
// Traverse each row of
// the given array
for (int i = 0; i < N; i++) {
// Initiliaze unique[]
// array to false
memset(unique, false,
sizeof(unique));
// Traverse each column
// of current row
for (int j = 0; j < N;
j++) {
// Stores the value
// of board[i][j]
int Z = board[i][j];
// Check if current row
// stores duplicate value
if (unique[Z]) {
return false;
}
unique[Z] = true;
}
}
// Traverse each column of
// the given array
for (int i = 0; i < N; i++) {
// Initiliaze unique[]
// array to false
memset(unique, false,
sizeof(unique));
// Traverse each row
// of current column
for (int j = 0; j < N;
j++) {
// Stores the value
// of board[j][i]
int Z = board[j][i];
// Check if current column
// stores duplicate value
if (unique[Z]) {
return false;
}
unique[Z] = true;
}
}
// Traverse each block of
// size 3 * 3 in board[][] array
for (int i = 0; i < N - 2;
i += 3) {
// j stores first column of
// each 3 * 3 block
for (int j = 0; j < N - 2;
j += 3) {
// Initiliaze unique[]
// array to false
memset(unique, false,
sizeof(unique));
// Traverse current block
for (int k = 0; k < 3;
k++) {
for (int l = 0; l < 3;
l++) {
// Stores row number
// of current block
int X = i + k;
// Stores column number
// of current block
int Y = j + l;
// Stores the value
// of board[X][Y]
int Z = board[X][Y];
// Check if current block
// stores duplicate value
if (unique[Z]) {
return false;
}
unique[Z] = true;
}
}
}
}
// If all conditions satisfied
return true;
}
// Driver Code
int main()
{
int board[N][N]
= { { 7, 9, 2, 1, 5, 4, 3, 8, 6 },
{ 6, 4, 3, 8, 2, 7, 1, 5, 9 },
{ 8, 5, 1, 3, 9, 6, 7, 2, 4 },
{ 2, 6, 5, 9, 7, 3, 8, 4, 1 },
{ 4, 8, 9, 5, 6, 1, 2, 7, 3 },
{ 3, 1, 7, 4, 8, 2, 9, 6, 5 },
{ 1, 3, 6, 7, 4, 8, 5, 9, 2 },
{ 9, 7, 4, 2, 1, 5, 6, 3, 8 },
{ 5, 2, 8, 6, 3, 9, 4, 1, 7 } };
if (isValidSudoku(board)) {
cout << "Valid";
}
else {
cout << "Not Valid";
}
} C#
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG{
static int N = 9;
// Function to check if all elements
// of the board[][] array store
// value in the range[1, 9]
static boolean isinRange(int[][] board)
{
// Traverse board[][] array
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
// Check if board[i][j]
// lies in the range
if (board[i][j] <= 0 ||
board[i][j] > 9)
{
return false;
}
}
}
return true;
}
// Function to check if the solution
// of sudoku puzzle is valid or not
static boolean isValidSudoku(int board[][])
{
// Check if all elements of board[][]
// stores value in the range[1, 9]
if (isinRange(board) == false)
{
return false;
}
// Stores unique value
// from 1 to N
boolean[] unique = new boolean[N + 1];
// Traverse each row of
// the given array
for(int i = 0; i < N; i++)
{
// Initiliaze unique[]
// array to false
Arrays.fill(unique, false);
// Traverse each column
// of current row
for(int j = 0; j < N; j++)
{
// Stores the value
// of board[i][j]
int Z = board[i][j];
// Check if current row
// stores duplicate value
if (unique[Z])
{
return false;
}
unique[Z] = true;
}
}
// Traverse each column of
// the given array
for(int i = 0; i < N; i++)
{
// Initiliaze unique[]
// array to false
Arrays.fill(unique, false);
// Traverse each row
// of current column
for(int j = 0; j < N; j++)
{
// Stores the value
// of board[j][i]
int Z = board[j][i];
// Check if current column
// stores duplicate value
if (unique[Z])
{
return false;
}
unique[Z] = true;
}
}
// Traverse each block of
// size 3 * 3 in board[][] array
for(int i = 0; i < N - 2; i += 3)
{
// j stores first column of
// each 3 * 3 block
for(int j = 0; j < N - 2; j += 3)
{
// Initiliaze unique[]
// array to false
Arrays.fill(unique, false);
// Traverse current block
for(int k = 0; k < 3; k++)
{
for(int l = 0; l < 3; l++)
{
// Stores row number
// of current block
int X = i + k;
// Stores column number
// of current block
int Y = j + l;
// Stores the value
// of board[X][Y]
int Z = board[X][Y];
// Check if current block
// stores duplicate value
if (unique[Z])
{
return false;
}
unique[Z] = true;
}
}
}
}
// If all conditions satisfied
return true;
}
// Driver Code
public static void main(String[] args)
{
int[][] board = { { 7, 9, 2, 1, 5, 4, 3, 8, 6 },
{ 6, 4, 3, 8, 2, 7, 1, 5, 9 },
{ 8, 5, 1, 3, 9, 6, 7, 2, 4 },
{ 2, 6, 5, 9, 7, 3, 8, 4, 1 },
{ 4, 8, 9, 5, 6, 1, 2, 7, 3 },
{ 3, 1, 7, 4, 8, 2, 9, 6, 5 },
{ 1, 3, 6, 7, 4, 8, 5, 9, 2 },
{ 9, 7, 4, 2, 1, 5, 6, 3, 8 },
{ 5, 2, 8, 6, 3, 9, 4, 1, 7 } };
if (isValidSudoku(board))
{
System.out.println("Valid");
}
else
{
System.out.println("Not Valid");
}
}
}
// This code is contributed by akhilsaini输出：
# Python3 program to implement
# the above approach
# Function to check if all elements
# of the board[][] array store
# value in the range[1, 9]
def isinRange(board):
N = 9
# Traverse board[][] array
for i in range(0, N):
for j in range(0, N):
# Check if board[i][j]
# lies in the range
if ((board[i][j] <= 0) or
(board[i][j] > 9)):
return False
return True
# Function to check if the solution
# of sudoku puzzle is valid or not
def isValidSudoku(board):
N = 9
# Check if all elements of board[][]
# stores value in the range[1, 9]
if (isinRange(board) == False):
return False
# Stores unique value
# from 1 to N
unique = [False] * (N + 1)
# Traverse each row of
# the given array
for i in range(0, N):
# Initiliaze unique[]
# array to false
for m in range(0, N + 1):
unique[m] = False
# Traverse each column
# of current row
for j in range(0, N):
# Stores the value
# of board[i][j]
Z = board[i][j]
# Check if current row
# stores duplicate value
if (unique[Z] == True):
return False
unique[Z] = True
# Traverse each column of
# the given array
for i in range(0, N):
# Initiliaze unique[]
# array to false
for m in range(0, N + 1):
unique[m] = False
# Traverse each row
# of current column
for j in range(0, N):
# Stores the value
# of board[j][i]
Z = board[j][i]
# Check if current column
# stores duplicate value
if (unique[Z] == True):
return False
unique[Z] = True
# Traverse each block of
# size 3 * 3 in board[][] array
for i in range(0, N - 2, 3):
# j stores first column of
# each 3 * 3 block
for j in range(0, N - 2, 3):
# Initiliaze unique[]
# array to false
for m in range(0, N + 1):
unique[m] = False
# Traverse current block
for k in range(0, 3):
for l in range(0, 3):
# Stores row number
# of current block
X = i + k
# Stores column number
# of current block
Y = j + l
# Stores the value
# of board[X][Y]
Z = board[X][Y]
# Check if current block
# stores duplicate value
if (unique[Z] == True):
return False
unique[Z] = True
# If all conditions satisfied
return True
# Driver Code
if __name__ == '__main__':
board = [ [ 7, 9, 2, 1, 5, 4, 3, 8, 6 ],
[ 6, 4, 3, 8, 2, 7, 1, 5, 9 ],
[ 8, 5, 1, 3, 9, 6, 7, 2, 4 ],
[ 2, 6, 5, 9, 7, 3, 8, 4, 1 ],
[ 4, 8, 9, 5, 6, 1, 2, 7, 3 ],
[ 3, 1, 7, 4, 8, 2, 9, 6, 5 ],
[ 1, 3, 6, 7, 4, 8, 5, 9, 2 ],
[ 9, 7, 4, 2, 1, 5, 6, 3, 8 ],
[ 5, 2, 8, 6, 3, 9, 4, 1, 7 ] ]
if (isValidSudoku(board)):
print("Valid")
else:
print("Not Valid")
# This code is contributed by akhilsaini时间复杂度： O(N 2 )
辅助空间： O(N)