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.."] ]
回溯法的思想。逐行选择所在的列，并判断是否满足条件，满足条件进入下一行。不满足，则返回。public List<String[]> solveNQueens(int n) { List <String[]> list = new ArrayList<String[]>(); int[] queenAtCol = new int[n]; getNQueens(0, n, queenAtCol, list); return list; } private void getNQueens(int row, int n, int[] queenAtCol, List<String[]> list) { if (row == n) { String[] str = new String[n]; for (int i = 0; i < n; i++) { StringBuilder builder = new StringBuilder(); for (int j = 0; j < n; j++) { if (queenAtCol[i] == j) { builder.append('Q'); } else { builder.append('.'); } } str[i] = builder.toString(); } list.add(str); } else { for (int col = 0; col < n; col++) { if (isValid(row,col,queenAtCol)) { queenAtCol[row] = col; getNQueens(row + 1, n, queenAtCol, list); } } } } private boolean isValid(int row, int col,int[] queenAtCol) { //row is valid. //column for (int i = 0; i < row; i++) { if (queenAtCol[i] == col) { return false; } } for (int i = 0; i < row; i++) { if (i + queenAtCol[i] == row + col) { return false; } } for (int i = 0; i < row; i++) { if (col - row == queenAtCol[i] - i) { return false; } } return true; }
原文地址：http://blog.csdn.net/u010378705/article/details/35553303
