Leetcode N-Queens I

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.."]]

写非递归回溯法要点：
1 记录好是否发生递归状态了
2 递归之后不能重复之前试过的解决方案
本题利用一个vector<int> backtrackCol(n,-1)实现了这两个功能
3 清楚什么时候需要回溯
本题是if (col == n) 这个条件就回溯了
4 回溯，就需要重置填过了的内容
本题需要重置两个数组： vector<int> backtrackCol和board
Iterative算法大概比recurrence算法难度高上三倍吧,而效率并没有明显区别。

下面是非递归算法12 queens用时也是12秒左右，和递归没什么区别。

可以打印的程序在下面连接：
http://blog.csdn.net/kenden23/article/details/14455915

vector<vector<string> > solveNQueens(int n) {vector<string> board(n,string(n, '.'));vector<vector<string> > rs;int row = 0;vector<int> backtrackCol(n, -1);while (row >= 0){int col = 0;if (backtrackCol[row] != -1){col = backtrackCol[row]+1;board[row][col-1] = '.';}for (; col < n; col++){if (isLegal(board, row, col)){board[row][col] = 'Q';backtrackCol[row] = col;if (row == n-1) {rs.push_back(board);backtrackCol[row] = -1;board[row][col] = '.';row--;break;}row++;break;}}if (col == n){backtrackCol[row] = -1;row--;}}return rs;}bool isLegal(const vector<string> &board, int row, int col){for (int i = 0; i < row; i++){if (board[i][col] == 'Q') return false;}for (int i = row - 1, j = col - 1; i>=0 && j>=0; i--, j--){if (board[i][j] == 'Q') return false;}for (int i = row - 1, j = col + 1; i>=0 && j<board.size(); i--, j++){if (board[i][j] == 'Q') return false;}return true;}

下面是递归回溯法，比上面简单多了。

vector<vector<string> > solveNQueens(int n) {vector<string> board(n,string(n, '.'));vector<vector<string> > rs;solNQueens(rs, board);return rs;}void solNQueens(vector<vector<string> > &rs, vector<string> &board, int row=0){if (row == board.size()) rs.push_back(board);for (int col = 0; col < board.size(); col++){if (isLegal(board, row, col)){board[row][col] = 'Q';solNQueens(rs, board, row+1);board[row][col] = '.';}}}bool isLegal(const vector<string> &board, int row, int col){for (int i = 0; i < row; i++){if (board[i][col] == 'Q') return false;}for (int i = row - 1, j = col - 1; i>=0 && j>=0; i--, j--){if (board[i][j] == 'Q') return false;}for (int i = row - 1, j = col + 1; i>=0 && j<board.size(); i--, j++){if (board[i][j] == 'Q') return false;}return true;}

递归回溯：

class Solution128 {public:vector<vector<string> > solveNQueens(int n) {vector<vector<string> > rs;vector<string> s(n, string(n, '.'));solve(rs, s, n);return rs;}void solve(vector<vector<string> > &rs, vector<string> &s, int n, int r = 0){if (r == n){rs.push_back(s);return;}for (int c = 0; c < n; c++){if (isLegal(s, r, c)){s[r][c] = 'Q';solve(rs, s, n, r+1);s[r][c] = '.';}}}bool isLegal(vector<string> &s, int r, int c){for (int i = r - 1, j = c - 1, k = c + 1; i >= 0 ; i--, j--, k++){if (s[i][c] == 'Q' || j>=0 && s[i][j] == 'Q' || k < s.size() && s[i][k] == 'Q')return false;}return true;}};