LeetCode-N-Queens 末蓝、 2022-08-04 10:43 140阅读 0赞 The n\-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other. ![8-queens.png][] 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,要保证任意两各queens都不能再同一行、同一列、同一对角线。代码如下: 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; } 其实这里面还是有很多值得研究的,首先搞了一个n长度的一维数组,而不是二维数组,因为数组的下标就代表了第几列,即ret\[col\] = row, 这是一个很好的优化。我上面是一列一列的扫描的,你也可以一行一行的扫描,都一样,因为这是个n\*n的正方形。 [8-queens.png]: http://www.leetcode.com/wp-content/uploads/2012/03/8-queens.png
还没有评论,来说两句吧...