LeetCode每日一题：n queens i

问题描述

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.."]
]

问题分析

n皇后问题，只要用递归和循环枚举就行了，n皇后属于回溯算法的一种。
计算是否在对角线只需要行的差和列的差的绝对值不要相等就可以了。

代码实现

 public ArrayList solveNQueens(int n) {
        ArrayList res = new ArrayList();
        helper1(n, 0, new int[n], res);
        return res;
    }

    private void helper1(int n, int row, int[] columnForRow, ArrayList res) {
        if (row == n) {
            String[] item = new String[n];
            for (int i = 0; i < n; i++) {
                StringBuilder strRow = new StringBuilder();
                for (int j = 0; j < n; j++) {
                    if (columnForRow[i] == j)
                        strRow.append('Q');
                    else
                        strRow.append('.');
                }
                item[i] = strRow.toString();
            }
            res.add(item);
            return;
        }
        for (int i = 0; i < n; i++) {
            columnForRow[row] = i;
            if (check(row, columnForRow)) {
                helper1(n, row + 1, columnForRow, res);
            }
        }
    }

    private boolean check(int row, int[] columnForRow) {
        for (int i = 0; i < row; i++) {
            if (columnForRow[row] == columnForRow[i] || Math.abs(columnForRow[row] - columnForRow[i]) == row - i)
                return false;
        }
        return true;
    }