原文:
给定一个整数 n ，任务是打印将 n 件放入 nxn 板的所有独特组合。
注意:打印(“*”)表示片段，打印(“-”)表示空白。
示例:
输入: n = 2 输出: * * –
– –
– –
––
– –
–– * 说明:空格总数为 22=4，要设置的棋子为 2，所以有 4c2 组合((4！/(2!*2!))=6)可能，如上所述。
输入:n = 1 t3】输出:*
方法:这个问题可以通过使用递归生成所有可能的解来解决。现在，按照以下步骤解决这个问题:
- 创建一个名为 allcombinations 的函数，它将生成所有可能的解。
- 它将取一个整数pieced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 表示当前件将要放置的行和列，以及一个字符串 ans 作为参数，用于存储放置件的矩阵。
- 现在，对所有组合的初始调用将通过 0 作为分段、 n 、 0 和 0 作为行和列，以及一个空字符串作为和。
- 在每次通话中，检查基本情况，即:
- 如果行变成 n 并且所有的片都被放置，即片被放置=n 。然后打印 ans 返回。否则如果件不是 n ，那么就从这个调用返回。
- 现在打两个电话:
- 一个是在当前位置加一个' ，一个是离开该位置加 '-'* 。
- 之后，递归调用将打印所有可能的pg电子试玩链接的解决方案。
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 语言(一种计算机语言，尤用于创建网站)
// 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, "");
}
}
python 3
# 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.
java 描述语言
// 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
output:
* *
- -
* -
* -
* -
- *
- *
* -
- *
- *
- -
* *
时间复杂度: o(2^m)，其中 m = n * n t3】辅助空间:t5】
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