LeetCode-N-Queens

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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。

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