leetcode -- N-Queens

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.."]
]
 1 public class Solution {
 2     public static void main(String[] args) {
 3         ArrayList<String[]> result = solveNQueens(4);
 4         System.out.println(result.size());
 5         for(int i = 0; i < result.size(); i++){
 6             System.out.println("solution:" + i);
 7             for(int j = 0; j < result.get(i).length; j++){
 8                 System.out.println(result.get(i)[j]);
 9             }
10         }
11     }
12 
13     public static ArrayList<String[]> solveNQueens(int n) {
14         // Start typing your Java solution below
15         // DO NOT write main() function
16         ArrayList<String[]> result = new ArrayList<String[]>();
17         int depth = 0;
18         int[] rows = new int[n];
19         for (int i = 0; i < n; i++) {
20             rows[i] = n + 1;
21         }
22         dfs(depth, rows, n, result);
23         return result;
24     }
25 
26     public static void dfs(int depth, int[] rows, int n, ArrayList<String[]> result) {
27         if (depth == n) {
28             String[] s = new String[n];
29             for (int i = 0; i < n; i++) {
30                 int m = rows[i];
31                 StringBuilder tmp = new StringBuilder();
32                 for (int j = 0; j < n; j++) {
33                     if (j == m) {
34                         tmp.append("Q");
35                         continue;
36                     }
37                     tmp.append(".");
38                 }
39                 s[i] = tmp.toString();
40             }
41             result.add(s);
42             return;
43         }
44         for (int i = 0; i < n; i++) {
45             rows[depth] = i;
46             if (isValid(rows, depth)) {
47                 dfs(depth + 1, rows, n, result);
48             }
49         }
50 
51     }
52     public static boolean isValid(int[] rows, int depth) {
53         for (int i = 0; i < depth; i++) {
54             if (rows[i] == rows[depth]
55                     || Math.abs(rows[i] - rows[depth]) == Math.abs(i - depth)) {
56                 return false;
57             }
58         }
59         return true;
60     }
61 }

 

posted @ 2013-08-11 14:18  feiling  阅读(367)  评论(0)  编辑  收藏  举报