【每日一题】- 2019-09-03 - 神奇的九位数 about leetcode HOT 5 CLOSED

/**
 *能被2整除的数，个位上的数能被2整除（偶数都能被2整除），那么这个数能被2整除

能被3整除的数，各个数位上的数字和能被3整除，那么这个数能被3整除

能被4整除的数，个位和十位所组成的两位数能被4整除，那么这个数能被4整除

能被5整除的数，个位上为0或5的数都能被5整除，那么这个数能被5整除

能被6整除的数，各数位上的数字和能被3整除的偶数，如果一个数既能被2整除又能被3整除，那么这个数能被6整除

能被7整除的数，若一个整数的个位数字截去，再从余下的数中，减去个位数的2倍，如果差是7的倍数，则原数能被7整除。如果差太大或心算不易看出是否7的倍数，就需要继续上述「截尾、倍大、相减、验差」的过程，直到能清楚判断为止。例如，判断133是否7的倍数的过程如下：13－3×2＝7，所以133是7的倍数；又例如判断6139是否7的倍数的过程如下：613－9×2＝595 ， 59－5×2＝49，所以6139是7的倍数，余类推。

能被8整除的数，一个整数的末3位若能被8整除，则该数一定能被8整除。

能被9整除的数，各个数位上的数字和能被9整除，那么这个数能被9整除
 *
 */
function find9Number() {
    for (var a = 1; a <= 9; a++) { //1
        for (var b = 1; b <= 9; b++) { // 2
            if (a == b) continue;
            if (b % 2 != 0) continue;

            for (var c = 1; c <= 9; c++) { //3
                if (c == b || c == a) {
                    continue;
                }
                if ((a + b + c) % 3 != 0) {
                    continue;
                }
                for (var d = 1; d <= 9; d++) { // 4
                    if (d == a || d == b || d == c) {
                        continue;
                    }

                    if ((c * 10 + d) % 4 != 0) continue;

                    for (var e = 1; e <= 9; e++) { // 5
                        if (e == a || e == b || e == c || e == d) {
                            continue;
                        }
                        if (e != 5) continue;

                        for (var f = 1; f <= 9; f++) { //6
                            if (f == a || f == b || f == c || f == d || f == e) {
                                continue;
                            }
                            if ((a + b + c + d + e + f) % 3 != 0 || f % 2 != 0) {
                                continue;
                            }
                            for (var g = 1; g <= 9; g++) { // 7
                                if (g == a || g == b || g == c || g == d || g == e || g == f) {
                                    continue;
                                }
                                var str = '' + a + b + c + d + e + f;
                                if ((Number(str) - g * 2) % 7 != 0) continue;

                                for (var h = 1; h <= 9; h++) { // 8
                                    if (h == a || h == b || h == c || h == d || h == e || h == f || h == g) continue;
                                    var str = '' + f + g + h;
                                    if (Number(str) % 8 != 0) continue;
                                    for (var i = 1; i <= 9; i++) { // 9
                                        if ((a + b + c + d + e + f + g + h + i) % 9 != 0) continue;

                                        console.log(a, b, c, d, e, f, g, h, i);
                                        return Number("" + a + b + c + d + e + f + g + h + i);
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}

function findNumber() {
    var flag = 1,
        map = new Set(),
        a = 1,
        b = 1,
        c = 1,
        d = 1,
        e = 1,
        f = 1,
        g = 1,
        h = 1,
        i = 1;
    while (true) {
        switch (flag) {
            case 1:
                while (map.has(a) && a <= 9) a++;
                a > 9 ? flag = 10 : (flag = 2, map.add(a));
                break;
            case 2:
                while ((map.has(b) || b % 2 != 0) && b <= 9) b++;
                b > 9 ? (flag = 1, map.delete(a), a++, b = 1) : (flag = 3, map.add(b));
                break;
            case 3:
                while ((map.has(c) || (a + b + c) % 3 != 0) && c <= 9) c++;
                c > 9 ? (flag = 2, map.delete(b), b++, c = 1) : (flag = 4, map.add(c));
                break;
            case 4:
                while ((map.has(d) || (c * 10 + d) % 4 != 0) && d <= 9) d++;
                d > 9 ? (flag = 3, map.delete(c), c++, d = 1) : (flag = 5, map.add(d));
                break;
            case 5:
                while ((map.has(e) || e != 5) && e <= 9) e++;
                e > 9 ? (flag = 4, map.delete(d), d++, e = 1) : (flag = 6, map.add(e));
                break;
            case 6:
                while ((map.has(f) || ((a + b + c + d + e + f) % 3 != 0 || f % 2 != 0)) && f <= 9) f++;
                f > 9 ? (flag = 5, map.delete(e), e++, f = 1) : (flag = 7, map.add(f));
                break;
            case 7:
                while ((map.has(g) || (Number("" + a + b + c + d + e + f) - g * 2) % 7 != 0) && g <= 9) g++;
                g > 9 ? (flag = 6, map.delete(f), f++, g = 1) : (flag = 8, map.add(g));
                break;
            case 8:
                while ((map.has(h) || Number('' + f + g + h) % 8 != 0) && h <= 9) h++;
                h > 9 ? (flag = 7, map.delete(g), g++, h = 1) : (flag = 9, map.add(h));
                break;
            case 9:
                while ((map.has(i) || (a + b + c + d + e + f + g + h + i) % 9 != 0) && i <= 9) i++;
                i > 9 ? (flag = 8, map.delete(h), h++, i = 1) : flag = 10;
                break;
            case 10:
                return "" + a + b + c + d + e + f + g + h + i;
        }
    }
}