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.."] ]八皇后问题的变形题,一个不能再经典的题了。
回溯法,要保证随意两各queens都不能再同一行、同一列、同一对角线。代码例如以下:
public List<List<String>> solveNQueens(int n) { List<List<String>> lists = new ArrayList<List<String>>(); dfs(lists, new int[n], 0); return lists; } private void dfs(List<List<String>> lists, int[] ret, int col) { if (col == ret.length) { List<String> list = new ArrayList<String>(ret.length); for (int p = 0; p < ret.length; p++) { StringBuilder sb = new StringBuilder(); for (int q = 0; q < ret.length; q++) { if (ret[p] == q) sb.append("Q"); else sb.append("."); } list.add(sb.toString()); } lists.add(list); return; } for (int k = 0; k < ret.length; k++) { ret[col] = k; if (check(ret, col)) dfs(lists, ret, col+1); } } private boolean check(int[] ret, int col) { for (int p = 0; p < col; p++) { if (Math.abs(ret[col]-ret[p]) == Math.abs(col-p)) return false; if (ret[col] == ret[p]) return false; } return true; }
事实上这里面还是有非常多值得研究的,首先搞了一个n长度的一维数组,而不是二维数组。由于数组的下标就代表了第几列。即ret[col] = row, 这是一个非常好的优化。我上面是一列一列的扫描的。你也能够一行一行的扫描。都一样。由于这是个n*n的正方形。