The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
Given an integer n, return all distinct solutions to the n-queens puzzle.
Each solution contains a distinct board configuration of the n-queens' placement, where 'Q'
and '.'
both
indicate a queen and an empty space respectively.
For example,
There exist two distinct solutions to the 4-queens puzzle:
[ [".Q..", // Solution 1 "...Q", "Q...", "..Q."], ["..Q.", // Solution 2 "Q...", "...Q", ".Q.."] ]
//N-QUEUE问题是深度优先搜索的很好运用,当然也可以用Stack来完成,但是深度优先搜索更加模式化,在面试中方便写代码。
public class Solution {
private String[] drawChessboard(ArrayList<Integer> cols) {
String[] chessboard = new String[cols.size()];
for (int i = 0; i < cols.size(); i++) {
chessboard[i] = "";
for (int j = 0; j < cols.size(); j++) {
if (j == cols.get(i)) {
chessboard[i] += "Q";
} else {
chessboard[i] += ".";
}
}
}
return chessboard;
}
//这个isValid的判断非常重要。
private boolean isValid(ArrayList<Integer> cols, int col) {
int row = cols.size();
for (int i = 0; i < row; i++) {
// same column
if (cols.get(i)== col) {
return false;
}
// left-top to right-bottom
// 代表如果是一个三角形,就说明两个Queue在一个对角线上,那么inValid
if (i - cols.get(i) == row - col) {
return false;
}
// right-top to left-bottom
if (i + cols.get(i) == row + col) {
return false;
}
}
return true;
}
//就是一个非常标准的深度优先搜索,不同的是不需要保存每个点是否访问过,因为isValid可以防止一些重复访问可能,但是大部分时间设置了比较好
private void search(int n, ArrayList<Integer> cols, ArrayList<String[]> result) {
if (cols.size() == n) {
result.add(drawChessboard(cols));
return;
}
for (int col = 0; col < n; col++) {
if (!isValid(cols, col)) {
continue;
}
cols.add(col);
search(n, cols, result);
cols.remove(cols.size() - 1);
}
}
public ArrayList<String[]> solveNQueens(int n) {
ArrayList<String[]> result = new ArrayList<String[]>();
if (n <= 0) {
return result;
}
search(n, new ArrayList<Integer>(), result);
return result;
}
}