1. 按每行只放一个皇后搜索

let n = 0;
let output = [];
//存放对已经放过皇后位置的标记
let row = [], col = [], diagonal = [], backDiagonal = [];

let init = n => {
    row = new Int32Array(n);
    col = new Int32Array(n);
    diagonal = new Int32Array(2 * n);
    backDiagonal = new Int32Array(2 * n);
    for (let i = 0; i < n; i++) {
        output[i] = new Array(n).fill('.');
    }
}

//按每行只能放一个进行搜索
let dfs = x => {
    if (x === n) { //每行都放完
        for (let i = 0; i < n; i++) {
            console.log(output[i].join(''));
        }
        console.log('');
        return;
    }
    for (let y = 0; y < n; y++) {
        // 检查每列和两对角线是否有冲突
        // (正对角线：y=-x+b;反对角线：y=x+b 则有 r=2,i=1与r=1,i=2 在同一条正对角线，反对角-i+n)
        if (col[y] || diagonal[x + y] || backDiagonal[x - y + n]) continue;
        col[y] = diagonal[x + y] = backDiagonal[x - y + n] = 1;
        output[x][y] = 'Q';
        dfs(x + 1);//进入下一行
        output[x][y] = '.';//回溯
        col[y] = diagonal[x + y] = backDiagonal[x - y + n] = 0;
    }
}

n = 4;
init(n);
dfs(0);

2. 按每个格子放皇后与不放皇后进行搜索

let n = 0;
let output = [];
//存放对已经放过皇后位置的标记
let row = [], col = [], diagonal = [], backDiagonal = [];

let init = n => {
    row = new Int32Array(n);
    col = new Int32Array(n);
    diagonal = new Int32Array(2 * n);
    backDiagonal = new Int32Array(2 * n);
    for (let i = 0; i < n; i++) {
        output[i] = new Array(n).fill('.');
    }
}

//按每个格子放皇后与不放皇后进行搜索
let dfs2 = (x, y, s) => {
    //从左到右，从上到下
    if (y === n) y = 0, x++;
    if (x === n) { //超出行数
        if (s === n) { //已放完
            for (let i = 0; i < n; i++) {
                console.log(output[i].join(''));
            }
            console.log('');
        }
        return;
    }
    // 分支1，不放皇后
    dfs2(x, y + 1, s);
    // 分支2，放皇后
    if (row[x] || col[y] || diagonal[x + y] || backDiagonal[x - y + n]) return;
    row[x] = col[y] = diagonal[x + y] = backDiagonal[x - y + n] = 1;
    output[x][y] = 'Q';
    dfs2(x, y + 1, s + 1);//进入下一个位置
    output[x][y] = '.';//回溯
    row[x] = col[y] = diagonal[x + y] = backDiagonal[x - y + n] = 0;
}

n = 4;
init(n);
dfs2(0, 0, 0);