您已获得数独谜题的解决方案。编写代码以检查它是否是有效的解决方案。
您的函数签名应该是：
boolean isValid(int starti, int startj, int endi, int endj)
不熟悉数独的规则：
【问题讨论】：
标签： sudoku
// rows
for (int i=0; i<9; i++) {
std::bitset<9> filled;
for (int j=0; j<9; j++)
filled.set(grid[i][j] - 1);
if (filled.count() != 9)
return false;
}
// ... similar with the loops "swapped" to get the columns
// (or do both in one loop)
for (int i=0; i<9; i += 3)
for (int j=0; j<9; j += 3) {
std::bitset<9> filled;
for (int k=0; k<3; k++)
for (int l=0; l<3; l++)
filled.set(grid[i+k][j+l] - 1);
if (filled.count() != 9)
return false;
}
return true;
【讨论】：
std::bitset（std::bitset.count 返回设置位数）检查每一行是否正好包含 9 个不同的数字。然后它以相同的方式检查每一列和每个块。
对不起，我知道这一定是家庭作业，但我忍不住。想出一些东西真是太有趣了:-)
一勺装满 LINQ 的药就下去了：
public class Sudoku
{
private int[][] sudoku;
public Sudoku(int[][] sudoku)
{
// TODO: Validate bounds and values
this.sudoku = sudoku;
}
public bool Validate() =>
VerticalLines.All(IsValid)
&& HorizontalLines.All(IsValid)
&& Squares.All(IsValid);
IEnumerable<IEnumerable<int>> VerticalLines =>
from line in sudoku select line;
IEnumerable<IEnumerable<int>> HorizontalLines =>
from y in Enumerable.Range(0, 9)
select (
from x in Enumerable.Range(0, 9)
select sudoku[x][y]);
IEnumerable<IEnumerable<int>> Squares =>
from x in Enumerable.Range(0, 3)
from y in Enumerable.Range(0, 3)
select GetSquare(x, y);
IEnumerable<int> GetSquare(int x, int y) =>
from squareX in Enumerable.Range(0, 3)
from squareY in Enumerable.Range(0, 3)
select sudoku[x * 3 + squareX][y * 3 + squareY];
bool IsValid(IEnumerable<int> line) => !(
from item in line
group item by item into g
where g.Count() > 1
select g)
.Any();
}
这个解决方案的好处是，如果你说你想出了这个，没有老师会相信你;-)
【讨论】：
SQL 允许您将规则定义为 CONSTRAINTS：无效的拼图被禁止并且不能存在。
-- Domain with only values 1..9 allowed,
-- used for both the {x,y,z} coordinates and the cell values.
CREATE DOMAIN one_nine
AS INTEGER
CHECK (value >= 1 AND value <= 9)
;
-- Table containing exactly one sudoku puzzle.
-- The zzz coordinate (the "box number") is formally redundant
-- (since it is functionally dependant on {xxx,yyy})
DROP TABLE IF EXISTS sudoku3 CASCADE;
CREATE TABLE sudoku3
( yyy one_nine NOT NULL
, xxx one_nine NOT NULL
, zzz one_nine NOT NULL
, val one_nine
);
-- First constraint: (x,y) is unique
ALTER TABLE sudoku3 ADD PRIMARY KEY (xxx,yyy);
-- Three constraints for unique values for {rows,columns,boxes}
CREATE UNIQUE INDEX sudoku_xv ON sudoku3 (xxx,val);
CREATE UNIQUE INDEX sudoku_yv ON sudoku3 (yyy,val);
CREATE UNIQUE INDEX sudoku_zv ON sudoku3 (zzz,val);
-- Create empty board.
INSERT INTO sudoku3 (yyy, xxx, zzz)
SELECT 1+(nn/9)
, 1+(nn%9)
, 1+3*((nn/3)%3)+1*(nn/27)
-- generate_series() is postgres-specific
FROM generate_series(0,80) AS nn;
现在，->val 成员可以更新，但生成的表永远不会违反四个约束中的任何一个（三个实际上是 INDEXES，但这在这里并不重要）。
检查一个完全有效的填充表意味着：检查是否有81个非NULL条目：
SELECT 1 AS result -- COUNT(*)
FROM sudoku3
WHERE val IS NOT NULL
HAVING COUNT(*) = 81
;
【讨论】：
使用位集的 Java 实现：
public static boolean isValid(int[][] board) {
//Check rows and columns
for (int i = 0; i < board.length; i++) {
BitSet bsRow = new BitSet(9);
BitSet bsColumn = new BitSet(9);
for (int j = 0; j < board[i].length; j++) {
if (board[i][j] == 0 || board[j][i] == 0) continue;
if (bsRow.get(board[i][j] - 1) || bsColumn.get(board[j][i] - 1))
return false;
else {
bsRow.set(board[i][j] - 1);
bsColumn.set(board[j][i] - 1);
}
}
}
//Check within 3 x 3 grid
for (int rowOffset = 0; rowOffset < 9; rowOffset += 3) {
for (int columnOffset = 0; columnOffset < 9; columnOffset += 3) {
BitSet threeByThree = new BitSet(9);
for (int i = rowOffset; i < rowOffset + 3; i++) {
for (int j = columnOffset; j < columnOffset + 3; j++) {
if (board[i][j] == 0) continue;
if (threeByThree.get(board[i][j] - 1))
return false;
else
threeByThree.set(board[i][j] - 1);
}
}
}
}
return true;
}
【讨论】：
这将是我在 ruby 中的解决方案
#Sudoku grid 9x9 with numbers between 1 and 9
#three rules:
#1) in any row all numbers between 1 and 9 have to be present
#2) in any columns all numbers between 1 and 9 have to be present
#3) in any of the 9 3x3 boxes all numbers between 1 and 9 have to be present
sudoku_correct =[[8,3,5,4,1,6,9,2,7],
[2,9,6,8,5,7,4,3,1],
[4,1,7,2,9,3,6,5,8],
[5,6,9,1,3,4,7,8,2],
[1,2,3,6,7,8,5,4,9],
[7,4,8,5,2,9,1,6,3],
[6,5,2,7,8,1,3,9,4],
[9,8,1,3,4,5,2,7,6],
[3,7,4,9,6,2,8,1,5]]
sudoku_incorrect =[[8,3,5,4,1,6,9,2,7],
[2,9,6,8,5,7,4,3,1],
[4,1,7,2,9,3,6,5,8],
[5,6,9,1,3,4,7,8,2],
[1,2,3,6,7,8,5,4,9],
[7,4,8,5,2,9,1,6,3],
[6,5,2,7,8,1,3,9,4],
[9,8,1,3,4,5,2,7,6],
[3,7,4,9,6,2,8,1,1]]
class Sudoku
def initialize(s_arr)
@sudoku_arr = s_arr
end
# given 9 integers makesure that you have 1 to 9
def valid_contents?(set)
set.each do |e|
return false unless (1..9).include?(e)
end
return true
end
# check if set has no duplicates
def has_no_duplicates?(set)
set.uniq.size < set.size ? false : true
end
def valid_set?(set)
valid_contents?(set) && has_no_duplicates?(set)
end
# obtain blocks of sudoku, given a grid
def get_blocks(s_grid)
blocks = []
s_grid.each_slice(3) do |row_set|
blocks_temp = [[],[],[]]
row_set.each do |row|
row.each_slice(3).with_index do|s,i|
blocks_temp[i] = blocks_temp[i] + s
end
end
blocks += blocks_temp
end
blocks
end
def valid?(s_arr = @sudoku_arr)
#check for row validity
s_arr.each do |set|
return false unless valid_set?(set)
end
#check for block validity
get_blocks(s_arr).each do |set|
return false unless valid_set?(set)
end
#check column validity
s_arr.transpose.each do |set|
return false unless valid_set?(set)
end
true
end
end
puts Sudoku.new(sudoku_correct).valid?
puts Sudoku.new(sudoku_incorrect).valid?
# output: True & False
【讨论】：
只使用一次数组迭代的解决方案。 通过数组的整个迭代，posRec 数组将在 ROW/COL/GRID 中具有每个数字的可用性。 如果 posRec 中缺少位，我们可以说数独不正确。
#include <stdio.h>
#define ROW 0
#define COL 1
#define GRID 2
#define TRUE 1
#define FALSE 0
char sudoku_correct[9][9] ={{8,3,5,4,1,6,9,2,7},
{2,9,6,8,5,7,4,3,1},
{4,1,7,2,9,3,6,5,8},
{5,6,9,1,3,4,7,8,2},
{1,2,3,6,7,8,5,4,9},
{7,4,8,5,2,9,1,6,3},
{6,5,2,7,8,1,3,9,4},
{9,8,1,3,4,5,2,7,6},
{3,7,4,9,6,2,8,1,5}};
char sudoku_incorrect[9][9] ={{8,3,5,4,1,6,9,2,7},
{2,9,6,8,5,7,4,3,1},
{4,1,7,2,9,3,6,5,8},
{5,6,9,1,3,4,7,8,2},
{1,2,3,6,7,8,5,4,9},
{7,4,8,5,2,9,1,6,3},
{6,5,2,7,8,1,3,9,4},
{9,8,1,3,4,5,2,7,6},
{3,7,4,9,6,2,8,1,1}};
short posRec[9][3];
int isValid(char a[][9]) {
int i, j, val, gIndex;
for(i=0; i <9; i++){
posRec[i][ROW] = 0;
posRec[i][COL] = 0;
posRec[i][GRID] = 0;
}
for(i=0; i<9; i++) {
for(j=0; j<9; j++) {
val=a[i][j]-1; //convert to 0-8 array index
if((posRec[val][ROW]>>i) & 0x1)
return FALSE;
posRec[val][ROW] |= (0x1<<i);
if((posRec[val][COL]>>j) & 0x1)
return FALSE;
posRec[val][COL] |= (0x1<<j);
gIndex = (j/3) + ((i/3) * 3);
if((posRec[val][GRID]>>gIndex) & 0x1)
return FALSE;
posRec[val][GRID] |= (0x1<<gIndex);
}
}
for(i=0; i<9;i++){
if(posRec[i][COL] != 0x1ff ||
posRec[i][ROW] != 0x1ff ||
posRec[i][GRID] != 0x1ff)
return FALSE;
}
return TRUE;
}
int main(){
printf("correct sudoku check = %s \n", isValid(sudoku_correct)?"CORRECT":"INCORRECT");
printf("incorrect sudoku check = %s \n", isValid(sudoku_incorrect)?"CORRECT":"INCORRECT");
return 0;
}
【讨论】：
package Questions;
public class SudukoSonChek {
public static void main(String args[]) {
int a[][] = { { 2, 4, 8, 3, 9, 5, 7, 1, 6 },
{ 5, 7, 1, 6, 2, 8, 3, 4, 9 }, { 9, 3, 6, 7, 4, 1, 5, 8, 2 },
{ 6, 8, 2, 5, 3, 9, 1, 7, 4 }, { 3, 5, 9, 1, 7, 4, 6, 2, 8 },
{ 7, 1, 4, 8, 6, 2, 9, 5, 3 }, { 8, 6, 3, 4, 1, 7, 2, 9, 5 },
{ 1, 9, 5, 2, 8, 6, 4, 3, 7 }, { 4, 2, 7, 9, 5, 3, 8, 6, 1 } };
System.out.println(check(a));
}
public static boolean check(int arr[][]) {
int i, j;
int count[] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 };
int count1[] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 };
boolean b = true;
for (i = 0; i < 9; i++) {
for (j = 0; j < 9; j++) {
if (count[arr[j][i]] > i) {
b = false;
return b;
}
if (count1[arr[i][j]] > i) {
b = false;
return b;
}
count1[arr[i][j]]++;
count[arr[j][i]]++;
}
}
return b;
}
}
【讨论】：
参考@Stephen 的出色回答，这里是 C# 中的一个解决方案，不使用 LINQ，以显示两种方法之间的区别。
此解决方案可处理 4x4 和 9x9 数独板。
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
namespace Sodoku
{
class Program
{
static void Main(string[] args)
{
List<string> lines = new List<string>();
lines.Add("4;1,4,2,3,2,3,1,4,4,2,3,1,3,1,4,2");
lines.Add("4;2,1,3,2,3,2,1,4,1,4,2,3,2,3,4,1");
lines.Add("9;5,3,4,6,7,8,9,1,2,6,7,2,1,9,5,3,4,8,1,9,8,3,4,2,5,6,7,8,5,9,7,6,1,4,2,3,4,2,6,8,5,3,7,9,1,7,1,3,9,2,4,8,5,6,9,6,1,5,3,7,2,8,4,2,8,7,4,1,9,6,3,5,3,4,5,2,8,6,1,7,9");
lines.Add("9;8,7,1,4,6,9,3,5,2,4,2,6,3,5,1,8,9,7,5,9,3,7,2,8,4,6,1,3,5,2,9,4,7,6,1,8,6,4,9,1,8,2,5,7,3,1,8,7,5,3,6,2,4,9,9,6,4,2,1,3,7,8,5,7,3,8,6,9,5,1,2,4,2,1,5,8,7,4,9,3,6");
lines.Add("9;1,2,7,5,3,9,8,4,6,4,5,3,8,6,1,7,9,2,8,9,6,4,7,2,1,5,3,2,8,9,3,1,7,4,6,5,3,6,5,2,8,4,9,1,7,7,4,1,9,5,6,3,2,8,9,7,4,6,2,8,5,3,1,5,1,2,7,4,3,6,8,9,6,3,8,1,9,5,2,7,4");
lines.Add("4;1,4,4,3,2,3,4,4,3,2,3,1,3,1,4,2");
foreach (var line in lines)
{
var tokens = line.Split(';');
var NxN = int.Parse(tokens[0]);
var nums = tokens[1].Trim().Split(',').Select(int.Parse).ToArray();
// Copy into Grid. Input is in row major form.
var grid = new int[NxN, NxN];
{
int i = 0;
for (int row = 0; row < NxN; row++)
{
for (int col = 0; col < NxN; col++)
{
grid[row, col] = nums[i];
i++;
}
}
}
int violations = 0;
// Check if each column passes tests for Sodoku.
{
for (int row = 0; row < NxN; row++)
{
var tempArray = new int[NxN];
for (int col = 0; col < NxN; col++)
{
tempArray[col] = grid[row, col];
}
if (IfArrayPassesSodoku(tempArray) == false)
{
violations++;
}
}
}
// Check if each row passes tests for Sodoku.
{
for (int row = 0; row < NxN; row++)
{
var tempArray = new int[NxN];
for (int col = 0; col < NxN; col++)
{
tempArray[col] = grid[col, row];
}
if (IfArrayPassesSodoku(tempArray) == false)
{
violations++;
}
}
}
// Check if each sub-grid passes tests for Sodoku.
// In 9x9 Sodoku, there are 9 subgrids of 3x3 each.
{
int NxNsub = (int)Math.Sqrt(NxN); // Length of side of sub-grid, for a 9x9 board this will be 3.
for (int row = 0; row < NxN; row += NxNsub)
{
for (int col = 0; col < NxN; col += NxNsub)
{
var tempArray = new int[NxN];
int index = 0;
for (int i = 0; i < NxNsub; i++)
{
for (int j = 0; j < NxNsub; j++)
{
tempArray[index] = grid[i + col, j + row];
index++;
}
}
if (IfArrayPassesSodoku(tempArray) == false)
{
violations++;
}
}
}
}
Console.WriteLine(violations == 0 ? "True" : "False");
// Correct output is:
// True
// False
// True
// True
// True
// False
}
Console.ReadKey();
}
/// <summary>
/// Checks to see if one-dimensional array passes Sodoku rules:
/// 1. Digits range from 1 to N.
/// 2. No repeating digits.
/// Could be way more efficient, but this solution is more readable compared to other concise forms.
/// </summary>
/// <param name="array">Array.</param>
/// <returns>True if one-dimensional array passes Sodoku rules.</returns>
static bool IfArrayPassesSodoku(int[] array)
{
int violations = 0;
// Check for repeating digits.
bool ifRepeatedDigits = (array.Distinct().ToArray().Length != array.Length);
if (ifRepeatedDigits == true)
{
return false;
}
// Check to see that it contains the digits 1 to N.
var sorted = array.OrderBy(o => o).ToArray();
if (array.Length == 4)
{
if (sorted[0] != 1 || sorted[3] != 4)
{
return false;
}
}
else if (array.Length == 9)
{
if (sorted[0] != 1 || sorted[8] != 9)
{
return false;
}
}
else
{
Console.Write("Error E5234. Invalid grid size.\n");
return false;
}
return true;
}
}
}
【讨论】：
我有一个类似的任务，输入来自标准输入。
我用过套装。 将元素放在正确的行、列和正方形（在我的代码中称为框）。 如果 3 个集合的大小都是 9，则有效。
#include <iostream>
#include <set>
int main()
{
//decl
std::set<int> row[9];
std::set<int> column[9];
std::set<int> box[9];
//read & store
for (int i = 0; i < 9; ++i)
{
for (int j = 0; j < 9; ++j)
{
int val;
std::cin >> val;
if ((val >= 1) && (val <= 9))
{
row[i].insert(val);
column[j].insert(val);
int k = j / 3 + i / 3 + i / 3 + i / 3;
box[k].insert(val);
}
}
}
//check
bool l = true;
int i = 0;
while (l && i < 9)
{
l = row[i].size() == 9;
++i;
}
if (!l) std::cout << "Invalid" << std::endl;
else
{
bool l = true;
int i = 0;
while (l && i < 9)
{
l = column[i].size() == 9;
++i;
}
if (!l) std::cout << "Invalid" << std::endl;
else
{
bool l = true;
int i = 0;
while (l && i < 9)
{
l = box[i].size() == 9;
++i;
}
if (!l) std::cout << "Invalid" << std::endl;
else std::cout << "Valid" << std::endl;
}
}
return 0;
}
【讨论】：
我认为我的解决方案很简洁，但如果其他人的解决方案显然也是如此。无论如何，我认为这是我在 2014 年 7 月 11 日为面试问题所做的一个非常清晰易读的 ruby 答案：
## Returns true iff the given string is a valid sudoku solution
def is_valid_solution(grid)
return false unless grid.is_a? String and grid.size == 9 * 9
validate_rows(grid) and validate_columns(grid) and validate_boxes(grid)
end
## Returns true iff the set is the chars "1", "2", ..., "9"
def is_valid_set(set)
return false if set.size != 9
(1..9).each do | x |
return false unless set.include?(x.to_s)
end
true
end
def validate_rows(grid)
(0...9).each do | row |
index = row * 9
return false unless is_valid_set(grid[index, 9])
end
true
end
def validate_columns(grid)
(0...9).each do | index |
column = (index...81).step(9).to_a.map { | i | grid[i] }.join
return false unless is_valid_set(column)
end
true
end
## This is ugly, but I'm running out of time...
TOP_LEFT_INDICES = [ 0, 1, 2, 9, 10, 11, 18, 19, 20]
TOP_MID_INDICES = [ 3, 4, 5, 12, 13, 14, 21, 22, 23]
TOP_RIGHT_INDICES = [ 6, 7, 8, 15, 16, 17, 24, 25, 26]
MID_LEFT_INDICES = [27, 28, 29, 36, 37, 38, 45, 46, 47]
MID_MID_INDICES = [30, 31, 32, 39, 40, 41, 48, 49, 50]
MID_RIGHT_INDICES = [33, 34, 35, 42, 43, 44, 51, 52, 53]
BOT_LEFT_INDICES = [54, 55, 56, 63, 64, 65, 72, 73, 74]
BOT_MID_INDICES = [57, 58, 59, 66, 67, 68, 75, 76, 77]
BOT_RIGHT_INDICES = [60, 61, 62, 69, 70, 71, 78, 79, 80]
BOX_INDICES = [TOP_LEFT_INDICES, TOP_MID_INDICES, TOP_RIGHT_INDICES,
MID_LEFT_INDICES, MID_MID_INDICES, MID_RIGHT_INDICES,
BOT_LEFT_INDICES, BOT_MID_INDICES, BOT_RIGHT_INDICES]
def validate_boxes(grid)
BOX_INDICES.each do | indices |
box = indices.map { | i | grid[i] }.join
return false unless is_valid_set(box)
end
true
end
【讨论】：
可以在单个循环中验证行和列
public class SudokuChecker {
static int[][] sMatrix = {
{5,3,4,6,7,8,9,1,2},
{6,7,2,1,9,5,3,4,8},
{1,9,8,3,4,2,5,6,7},
{8,5,9,7,6,1,4,2,3},
{4,2,6,8,5,3,7,9,1},
{7,1,3,9,2,4,8,5,6},
{9,6,1,5,3,7,2,8,4},
{2,8,7,4,1,9,6,3,5},
{3,4,5,2,8,6,1,7,9}
};
public static void main(String[] args) {
if (rowColumnCheck() && boxCheck()) {
System.out.println("Valid!");
} else {
System.out.println("Invalid!");
}
}
private static boolean boxCheck() {
for (int i = 0; i < sMatrix.length; i += 3) {
for (int j = 0; j < sMatrix.length; j += 3) {
boolean[] gridMatrix = new boolean[9];
for (int k = i; k < 3 + i; k++) {
for (int l = j; l < 3 + j; l++) {
int currentNumber = sMatrix[k][l];
if (currentNumber < 1 || currentNumber > 9) {
return false;
}
gridMatrix[currentNumber - 1] = true;
}
}
for (boolean booleanValue : gridMatrix) {
if (!booleanValue) {
return false;
}
}
}
}
return true;
}
private static boolean rowColumnCheck() {
for (int i = 0; i < 9; i++) {
boolean[] rowArray = new boolean[9];
boolean[] columnArray = new boolean[9];
for (int j = 0; j < 9; j++) {
int currentNumberRow = sMatrix[i][j];
int currentNumberColumn = sMatrix[j][i];
if ((currentNumberRow < 1 || currentNumberRow > 9) && (currentNumberColumn < 1 || currentNumberColumn > 9)) {
return false;
}
rowArray[currentNumberRow - 1] = true;
columnArray[currentNumberColumn - 1] = true;
}
for (boolean booleanValue : rowArray) {
if (!booleanValue) {
return false;
}
}
for (boolean booleanValue : columnArray) {
if (!booleanValue) {
return false;
}
}
}
return true;
}
}
【讨论】：
sum = 45 是否有效？如果列有37, 1, 1, 1, 1, 1, 1, 1, 1 怎么办？你的逻辑也有缺陷——如果第一行是正确的successTillNow 将是true——如果第二行是错误的，它将返回successTillNow 的当前值，它仍然是true！