原题链接:https://leetcode.com/problems/n-queens/description/
这道题目就是由鼎鼎大名的八皇后问题延伸而来的 n 皇后问题,我看的《数据结构(C语言版)》上面树章节里面也提到了这个问题,说是使用典型的回溯法。这道题目本身我是没有想出解法的,官方也无解答,还是看了讨论区评分最高的答案。答案不好理解,还是自己放到 IDE 里面调试运行下就理解大致思路了:
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
/**
* Created by clearbug on 2018/2/26.
*/
public class Solution {
public static void main(String[] args) {
Solution s = new Solution();
List<List<String>> res4 = s.solveNQueens(4);
List<List<String>> res8 = s.solveNQueens(8);
System.out.println(s.solveNQueens(4));
System.out.println(s.solveNQueens(8));
}
public List<List<String>> solveNQueens(int n) {
char[][] board = new char[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
board[i][j] = ‘.‘;
}
}
List<List<String>> res = new ArrayList<>();
dfs(board, 0, res);
return res;
}
private void dfs(char[][] board, int colIndex, List<List<String>> res) {
if (colIndex == board.length) {
res.add(construct(board));
return;
}
for (int i = 0; i < board.length; i++) {
if (validate(board, i, colIndex)) {
board[i][colIndex] = ‘Q‘;
dfs(board, colIndex + 1, res);
board[i][colIndex] = ‘.‘;
}
}
}
private boolean validate(char[][] board, int x, int y) {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < y; j++) {
if (board[i][j] == ‘Q‘ && (x + j == y + i || x + y == i + j || x == i)) {
return false;
}
}
}
return true;
}
private List<String> construct(char[][] board) {
List<String> res = new LinkedList<>();
for (int i = 0; i < board.length; i++) {
String s = new String(board[i]);
res.add(s);
}
return res;
}
}
这位老铁的答案算是勉强看懂了,百度百科的答案也不知道是谁贴的,真没看懂。。。
参考
https://baike.baidu.com/item/%E5%85%AB%E7%9A%87%E5%90%8E%E9%97%AE%E9%A2%98/11053477?fr=aladdin