在 NxN 板上打印设置 N 件的所有独特组合
给定一个整数N ，任务是打印将N块放在NxN板上的所有唯一组合。
注意：打印（“*”）表示碎片，（“-”）表示空白。
例子：
Input: N = 2
Output:
* *
– –
* –
* –
* –
– *
– *
* –
– *
– *
– –
* *
Explanation: The total number of empty spaces are 2*2=4 and the pieces to be set is 2 so there are 4C2 combinations ((4!/(2!*2!))=6) possible which is represented above.
Input: N = 1
Output: *
方法：这个问题可以通过使用递归生成所有可能的解决方案来解决。现在，请按照以下步骤解决此问题：
- 创建一个名为allCombinations的函数，它将生成所有可能的解决方案。
- 它将采用一个整数piecesPlaced表示放置的总件数，整数N表示需要放置的件数，两个整数row和col表示将要放置当前件的行和列，以及一个字符串ans for存储放置片段的矩阵作为参数。
- 现在，对allCombinations的初始调用将传递0作为piecesPlaced ， N 、 0和0作为row和col以及一个空字符串作为ans 。
- 在每次调用中，检查基本情况，即：
- 如果 row 变为N并且所有棋子都已放置，即piecesPlaced=N 。然后打印答案并返回。否则，如果piecesPlaced不是N ，则只需从此调用返回。
- 现在打两个电话：
- 一种是在当前位置添加一个'*' ，另一种是离开该位置并添加'-' 。
- 在此之后，递归调用将打印所有可能的解决方案。
下面是上述方法的实现。
C++
// C++ Program for the above approach
#include
using namespace std;
// Function to print all
// combinations of setting N
// pieces in N x N board
void allCombinations(int piecesPlaced, int N, int row,
int col, string ans)
{
// If the total 2d array's space
// is exhausted then backtrack.
if (row == N) {
// If all the pieces are
// placed then print the answer.
if (piecesPlaced == N) {
cout << ans;
}
return;
}
int nr = 0;
int nc = 0;
// Declare one string
// that will set the piece.
string x = "";
// Declare one string that
// will leave the space blank.
string y = "";
// If the current column
// is out of bounds then
// increase the row
// and set col to 0.
if (col == N - 1) {
nr = row + 1;
nc = 0;
x = ans + "*\n";
y = ans + "-\n";
}
// Else increase the col
else {
nr = row;
nc = col + 1;
x = ans + "*\t";
y = ans + "-\t";
}
// Set the piece in the
// box and move ahead
allCombinations(piecesPlaced + 1, N, nr, nc, x);
// Leave the space blank
// and move forward
allCombinations(piecesPlaced, N, nr, nc, y);
}
// Driver Code
int main()
{
int N = 2;
allCombinations(0, N, 0, 0, "");
return 0;
}
// This code is contributed by rakeshsahni. Java
// Java Program for the above approach
import java.io.*;
import java.util.*;
public class main {
// Function to print all
// combinations of setting N
// pieces in N x N board
public static void allCombinations(
int piecesPlaced,
int N, int row,
int col, String ans)
{
// If the total 2d array's space
// is exhausted then backtrack.
if (row == N) {
// If all the pieces are
// placed then print the answer.
if (piecesPlaced == N) {
System.out.println(ans);
}
return;
}
int nr = 0;
int nc = 0;
// Declare one string
// that will set the piece.
String x = "";
// Declare one string that
// will leave the space blank.
String y = "";
// If the current column
// is out of bounds then
// increase the row
// and set col to 0.
if (col == N - 1) {
nr = row + 1;
nc = 0;
x = ans + "*\n";
y = ans + "-\n";
}
// Else increase the col
else {
nr = row;
nc = col + 1;
x = ans + "*\t";
y = ans + "-\t";
}
// Set the piece in the
// box and move ahead
allCombinations(
piecesPlaced + 1, N,
nr, nc, x);
// Leave the space blank
// and move forward
allCombinations(piecesPlaced, N,
nr, nc, y);
}
// Driver Code
public static void main(String[] args)
throws Exception
{
int N = 2;
allCombinations(0, N, 0, 0, "");
}
}Python3
# Python Program for the above approach
# Function to print all
# combinations of setting N
# pieces in N x N board
def allCombinations(piecesPlaced, N, row, col, ans):
# If the total 2d array's space
# is exhausted then backtrack.
if row == N:
# If all the pieces are
# placed then print the answer.
if piecesPlaced == N:
print(ans)
return;
nr = 0
nc = 0
# Declare one string
# that will set the piece.
x = ""
# Declare one string that
# will leave the space blank.
y = ""
# If the current column
# is out of bounds then
# increase the row
# and set col to 0.
if col == N - 1:
nr = row + 1
nc = 0
x = ans + "*\n"
y = ans + "-\n"
# Else increase the col
else:
nr = row
nc = col + 1
x = ans + "* "
y = ans + "- "
# Set the piece in the
# box and move ahead
allCombinations(piecesPlaced + 1, N, nr, nc, x);
# Leave the space blank
# and move forward
allCombinations(piecesPlaced, N, nr, nc, y);
# Driver Code
N = 2
allCombinations(0, N, 0, 0, "")
# This code is contributed by rdtank.C#
// C# Program for the above approach
using System;
public class main {
// Function to print all
// combinations of setting N
// pieces in N x N board
public static void allCombinations(int piecesPlaced,
int N, int row,
int col, String ans)
{
// If the total 2d array's space
// is exhausted then backtrack.
if (row == N) {
// If all the pieces are
// placed then print the answer.
if (piecesPlaced == N) {
Console.WriteLine(ans);
}
return;
}
int nr = 0;
int nc = 0;
// Declare one string
// that will set the piece.
String x = "";
// Declare one string that
// will leave the space blank.
String y = "";
// If the current column
// is out of bounds then
// increase the row
// and set col to 0.
if (col == N - 1) {
nr = row + 1;
nc = 0;
x = ans + "*\n";
y = ans + "-\n";
}
// Else increase the col
else {
nr = row;
nc = col + 1;
x = ans + "*\t";
y = ans + "-\t";
}
// Set the piece in the
// box and move ahead
allCombinations(piecesPlaced + 1, N, nr, nc, x);
// Leave the space blank
// and move forward
allCombinations(piecesPlaced, N, nr, nc, y);
}
// Driver Code
public static void Main(string[] args)
{
int N = 2;
allCombinations(0, N, 0, 0, "");
}
}
// This code is contributed by ukasp.Javascript
// Javascript Program for the above approach
// Function to print all
// combinations of setting N
// pieces in N x N board
function allCombinations(piecesPlaced, N, row, col, ans) {
// If the total 2d array's space
// is exhausted then backtrack.
if (row == N) {
// If all the pieces are
// placed then print the answer.
if (piecesPlaced == N) {
document.write(ans);
}
return;
}
let nr = 0;
let nc = 0;
// Declare one string
// that will set the piece.
let x = "";
// Declare one string that
// will leave the space blank.
let y = "";
// If the current column
// is out of bounds then
// increase the row
// and set col to 0.
if (col == N - 1) {
nr = row + 1;
nc = 0;
x = ans + "*
";
y = ans + "-
";
}
// Else increase the col
else {
nr = row;
nc = col + 1;
x = ans + "* ";
y = ans + "- ";
}
// Set the piece in the
// box and move ahead
allCombinations(piecesPlaced + 1, N, nr, nc, x);
// Leave the space blank
// and move forward
allCombinations(piecesPlaced, N, nr, nc, y);
}
// Driver Code
let N = 2;
allCombinations(0, N, 0, 0, "");
// This code is contributed by Saurabh Jaiswal输出：
* *
- -
* -
* -
* -
- *
- *
* -
- *
- *
- -
* *时间复杂度： O(2^M)，其中 M=N*N
辅助空间： 