N-Queens I&&II

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.."]
]

深搜可行解，对于每个皇后，它所在的行、列、两条斜方向都必须没有其他的皇后，这样才不会引起冲突，那么我们就对棋盘每行进行搜索，对每一行可以放皇后的的格子均深搜，如果能搜到n行，那么说明目前棋盘的布局是N皇后解的一种

 1 class Solution {
 2 public:
 3     vector<vector<string> > solveNQueens(int n) {
 4         ans.clear();
 5         board.resize(n);    //初始化棋盘
 6         for(int i=0; i<n; ++i)
 7             board[i].resize(n, false);
 8         dfs(0, n);
 9         return ans;
10     }
11
12     void dfs(int k, int n) {
13         if( k==n ) {
14             vector<string> vstr = board2str();
15             ans.push_back(vstr);
16             return ;
17         }
18         for(int i=0; i<n; ++i)
19             if( canPut(k, i, n) ) {
20                 board[k][i] = true;
21                 dfs(k+1, n);
22                 board[k][i] = false;
23             }
24     }
25
26     bool canPut(int x, int y, int n) {
27         for(int i=0; i<n; ++i)  //横向
28             if( board[x][i] ) return false;
29         for(int i=0; i<n; ++i)  //纵向
30             if( board[i][y] ) return false;
31         int i = x;
32         int j = y;
33         //斜左向上
34         while( i>=0 && j>=0 ) {
35             if( board[i][j] ) return false;
36             --i,--j;
37         }
38         //斜右向下
39         i=x, j=y;
40         while( i<n && j<n ) {
41             if( board[i][j] ) return false;
42             ++i,++j;
43         }
44         //斜右向上
45         i=x, j=y;
46         while( i>=0 && j<n ) {
47             if( board[i][j] ) return false;
48             --i,++j;
49         }
50         //斜左向下
51         i=x, j=y;
52         while( i<n && j>=0 ) {
53             if( board[i][j] ) return false;
54             ++i,--j;
55         }
56         return true;
57     }
58
59     vector<string> board2str() {    //将棋盘转为string
60         vector<string> ans;
61         for(int i=0; i<board.size(); ++i) {
62             string str;
63             for(int j=0; j<board[i].size(); ++j)
64                 str.push_back( board[i][j] ? ‘Q‘ : ‘.‘ );
65             ans.push_back(str);
66         }
67         return ans;
68     }
69
70 private:
71     vector< vector<bool> > board;
72     vector< vector<string> > ans;
73 };

N-Queens II

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

改动I中所用的方法，直接获取解的个数。