Difficulty: Hard
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.
Example:
Input: 4
Output: [
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],
["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
Language: Java
class Solution {
public List<List<String>> solveNQueens(int n) {
List<List<String>> result = new ArrayList<>();
if (n <= 0) {
return result;
}
LinkedList<Integer> column = new LinkedList<>();
LinkedList<Integer> pie = new LinkedList<>();
LinkedList<Integer> na = new LinkedList<>();
this.DFS(column, pie, na,
0, n, result, new ArrayList<>());
return result;
}
private void DFS(LinkedList<Integer> column, LinkedList<Integer> pie, LinkedList<Integer> na,
int row, int n, List<List<String>> result, List<String> currStat) {
if (row >= n) {
result.add(currStat);
return;
}
for (int i = 0; i < n; i++) {
if (column.contains(i) || pie.contains(i + row) || na.contains(row - i)) {
continue;
}
// 记录皇后可以攻击的点
column.add(i);
pie.add(i + row);
na.add(row - i);
StringBuilder stringBuilder = new StringBuilder();
for (int j = 0; j < n; j++) {
if (j == i) {
stringBuilder.append("Q");
} else {
stringBuilder.append(".");
}
}
this.DFS(column, pie, na, row + 1, n, result, new ArrayList<String>(currStat) {{
add(stringBuilder.toString());
}});
// 移除，便于下一次回溯
column.pollLast();
pie.pollLast();
na.pollLast();
}
}
}