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algorithmzuoalgorithmzuo
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‎src/class084/Code01_CountNumbersWithUniqueDigits.java‎

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package class084;
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// 统计各位数字都不同的数字个数
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// 给你一个整数n，代表十进制数字最多有n位
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// 如果某个数字，每一位都不同，那么这个数字叫做有效数字
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// 返回有效数字的个数，不统计负数范围
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// 测试链接 : https://leetcode.cn/problems/count-numbers-with-unique-digits/
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public class Code01_CountNumbersWithUniqueDigits {
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public static int countNumbersWithUniqueDigits(int n) {
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if (n == 0) {
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return 1;
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}
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int ans = 10;
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for (int s = 9, i = 9, k = 2; k <= n; i--, k++) {
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s *= i;
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ans += s;
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}
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return ans;
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}
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}

‎src/class084/Code02_NumbersWithRepeatedDigits.java‎

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package class084;
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// 至少有1位重复的数字
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// 给定正整数n
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// 返回在[1, n]范围内具有至少1位重复数字的正整数的个数
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// 测试链接 : https://leetcode.cn/problems/numbers-with-repeated-digits/
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public class Code02_NumbersWithRepeatedDigits {
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public static int numDupDigitsAtMostN(int n) {
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if (n <= 10) {
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return 0;
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}
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int len = 1;
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int offset = 1;
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int tmp = n / 10;
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while (tmp > 0) {
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len++;
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offset *= 10;
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tmp /= 10;
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}
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int noRepeat = 0;
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for (int i = 1, cnt = 9, cur = 9; i < len; i++) {
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if (i == 1) {
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noRepeat += 10;
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} else {
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cnt *= cur--;
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noRepeat += cnt;
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}
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}
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int[] cnt = new int[len];
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cnt[0] = 1;
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for (int i = 1, ans = 1, base = 10 - len + 1; i < len; i++, base++) {
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ans *= base;
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cnt[i] = ans;
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}
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int status = 0b1111111111;
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noRepeat += ((n / offset) - 1) * cnt[len - 1];
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noRepeat += f(cnt, len - 1, offset / 10, status ^ (1 << (n / offset)), n);
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return n + 1 - noRepeat;
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}
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public static int f(int[] cnt, int len, int offset, int status, int n) {
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if (len == 0) {
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return 1;
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}
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int ans = 0;
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int first = (n / offset) % 10;
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for (int cur = 0; cur < first; cur++) {
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if ((status & (1 << cur)) != 0) {
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ans += cnt[len - 1];
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}
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}
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if ((status & (1 << first)) != 0) {
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ans += f(cnt, len - 1, offset / 10, status ^ (1 << first), n);
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}
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return ans;
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}
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}

‎src/class084/Code03_NumbersAtMostGivenDigitSet.java‎

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package class084;
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// 最大为N的数字组合
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// 给定一个按 非递减顺序 排列的数字数组 digits
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// 你可以用任意次数 digits[i] 来写的数字
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// 例如，如果 digits = ['1','3','5']
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// 我们可以写数字，如 '13', '551', 和 '1351315'
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// 返回 可以生成的小于或等于给定整数 n 的正整数的个数
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// 测试链接 : https://leetcode.cn/problems/numbers-at-most-n-given-digit-set/
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public class Code03_NumbersAtMostGivenDigitSet {
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public static int atMostNGivenDigitSet(String[] strs, int number) {
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int m = strs.length;
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int[] digits = new int[m];
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for (int i = 0; i < m; i++) {
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digits[i] = Integer.valueOf(strs[i]);
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}
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int tmp = number / 10;
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int len = 1;
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int offset = 1;
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while (tmp > 0) {
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tmp /= 10;
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len++;
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offset *= 10;
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}
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int[] cnt = new int[len];
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cnt[0] = 1;
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int ans = 0;
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for (int prepare = m, i = 1; i < len; i++, prepare *= m) {
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cnt[i] = prepare;
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ans += prepare;
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}
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return ans + f(digits, cnt, number, offset, len);
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}
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public static int f(int[] digits, int[] cnt, int number, int offset, int len) {
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if (len == 0) {
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return 1;
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}
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int cur = (number / offset) % 10;
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int ans = 0;
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for (int i = 0; i < digits.length; i++) {
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if (digits[i] < cur) {
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ans += cnt[len - 1];
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} else if (digits[i] == cur) {
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ans += f(digits, cnt, number, offset / 10, len - 1);
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} else {
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break;
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}
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}
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return ans;
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}
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}

‎src/class084/Code04_CountSpecialIntegers.java‎

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package class084;
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// 统计特殊整数
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// 如果一个正整数每一个数位都是 互不相同 的，我们称它是 特殊整数
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// 给你一个正整数n，请你返回区间 [1, n] 之间特殊整数的数目
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// 测试链接 : https://leetcode.cn/problems/count-special-integers/
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public class Code04_CountSpecialIntegers {
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public static int[] offset = {
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0, // 0
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1, // 1
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10, // 2
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100, // 3
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1000, // 4
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10000, // 5
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100000, // 6
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1000000, // 7
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10000000, // 8
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100000000, // 9
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1000000000 // 10
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};
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public static int countSpecialNumbers(int n) {
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int len = len(n);
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int ans = 0;
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for (int i = 1; i < len; i++) {
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ans += all(i);
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}
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int firstNumber = n / offset[len];
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ans += (firstNumber - 1) * small(len - 1, 9);
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ans += f(n, len, len - 1, 1 << firstNumber);
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return ans;
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}
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public static int len(int n) {
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int ans = 0;
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while (n != 0) {
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ans++;
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n /= 10;
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}
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return ans;
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}
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public static int all(int bits) {
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int ans = 9;
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int cur = 9;
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while (--bits != 0) {
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ans *= cur--;
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}
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return ans;
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}
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public static int small(int bits, int candidates) {
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int ans = 1;
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for (int i = 0; i < bits; i++, candidates--) {
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ans *= candidates;
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}
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return ans;
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}
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public static int f(int num, int len, int rest, int status) {
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if (rest == 0) {
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return 1;
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}
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int cur = (num / offset[rest]) % 10;
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int cnt = 0;
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for (int i = 0; i < cur; i++) {
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if ((status & (1 << i)) == 0) {
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cnt++;
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}
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}
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int ans = cnt * small(rest - 1, 9 - (len - rest));
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if ((status & (1 << cur)) == 0) {
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ans += f(num, len, rest - 1, status | (1 << cur));
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}
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return ans;
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}
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}

‎src/class085/Code01_CountOfIntegers.java‎

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package class085;
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// 统计整数数目
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// 给你两个数字字符串 num1 和 num2 ，以及两个整数max_sum和min_sum
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// 如果一个整数 x 满足以下条件，我们称它是一个好整数
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// num1 <= x <= num2
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// min_sum <= digit_sum(x) <= max_sum
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// 请你返回好整数的数目
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// 答案可能很大请返回答案对10^9 + 7 取余后的结果
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// 注意，digit_sum(x)表示x各位数字之和
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// 测试链接 : https://leetcode.cn/problems/count-of-integers/
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public class Code01_CountOfIntegers {
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public static int MOD = 1000000007;
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public static int MAXN = 23;
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public static int MAXM = 401;
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public static int[][][] dp = new int[MAXN][MAXM][2];
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public static void build() {
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for (int i = 0; i < n; i++) {
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for (int j = 0; j <= m; j++) {
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dp[i][j][0] = -1;
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dp[i][j][1] = -1;
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}
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}
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}
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public static char[] num;
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public static int n, m;
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public static int count(String num1, String num2, int min_sum, int max_sum) {
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num = num2.toCharArray();
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n = num2.length();
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m = max_sum;
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build();
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int ans = f(0, 0, 0, min_sum, max_sum);
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num = num1.toCharArray();
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n = num1.length();
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build();
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ans = (ans - f(0, 0, 0, min_sum, max_sum) + MOD) % MOD;
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if (check(min_sum, max_sum)) {
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ans = (ans + 1) % MOD;
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}
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return ans;
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}
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public static int f(int i, int sum, int less, int min, int max) {
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if (sum > max) {
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return 0;
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}
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if (sum + (n - i) * 9 < min) {
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return 0;
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}
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if (i == n) {
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return 1;
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}
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if (dp[i][sum][less] != -1) {
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return dp[i][sum][less];
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}
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int cur = num[i] - '0';
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int ans = 0;
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if (less == 0) {
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for (int pick = 0; pick < cur; pick++) {
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ans = (ans + f(i + 1, sum + pick, 1, min, max)) % MOD;
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}
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ans = (ans + f(i + 1, sum + cur, 0, min, max)) % MOD;
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} else {
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for (int pick = 0; pick <= 9; pick++) {
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ans = (ans + f(i + 1, sum + pick, 1, min, max)) % MOD;
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}
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}
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dp[i][sum][less] = ans;
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return ans;
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}
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public static boolean check(int min, int max) {
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int sum = 0;
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for (char cha : num) {
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sum += cha - '0';
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}
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return sum >= min && sum <= max;
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}
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}

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