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.."]
]
Have you met this question in a real interview?
 
Analysis:
How to determine whether the current position at jth row is valid? Suppose jth row position is place[j] and ith row position is place[i]. Then we found that place[i] leads three positions at jth row invalid. They are place[i], place[i]+j-i, place[i]-(j-i).
 
Solution:
 1 public class Solution {
 2     public List<String[]> solveNQueens(int n) {
 3         List<String[]> res = new ArrayList<String[]>();
 4         int[] place = new int[n];
 5         Arrays.fill(place,-1);
 6         int level = 0;
 7   
 8         while (level!=-1){
 9             if (level>=n){
10                 constructRes(place,res,n);
11                 level--;
12                 continue;
13             }
14 
15             int val = place[level];
16             val++;
17             while (val<n){
18                 boolean valid = true;
19                 for (int i=0;i<level;i++)
20                     if (val==place[i] || val==place[i]+level-i || val==place[i]-(level-i)){
21                         valid = false;
22                         break;
23                     }
24                 if (valid) break;
25                 else val++;
26             }
27 
28             if (val<n){
29                 place[level]=val;
30                 level++;
31             } else {
32                 place[level]=-1;
33                 level--;
34             }
35         }
36 
37         return res;
38         
39     }
40 
41     public void constructRes(int[] place, List<String[]> res, int n){
42         char[] line = new char[n];
43         Arrays.fill(line,'.');
44         String[] oneRes = new String[n];
45         for (int i=0;i<n;i++){
46             int val = place[i];
47             line[val]='Q';
48             String lineStr = new String(line);
49             line[val]='.';
50             oneRes[i]=lineStr;
51         }
52         res.add(oneRes);
53     }       
54 }

 

 
 
posted @ 2014-11-26 22:55  LiBlog  阅读(139)  评论(0)  编辑  收藏  举报