题目：

题解：

class Solution {public int strongPasswordChecker(String password) {int n = password.length();int hasLower = 0, hasUpper = 0, hasDigit = 0;for (int i = 0; i < n; ++i) {char ch = password.charAt(i);if (Character.isLowerCase(ch)) {hasLower = 1;} else if (Character.isUpperCase(ch)) {hasUpper = 1;} else if (Character.isDigit(ch)) {hasDigit = 1;}}int categories = hasLower + hasUpper + hasDigit;if (n < 6) {return Math.max(6 - n, 3 - categories);} else if (n <= 20) {int replace = 0;int cnt = 0;char cur = '#';for (int i = 0; i < n; ++i) {char ch = password.charAt(i);if (ch == cur) {++cnt;} else {replace += cnt / 3;cnt = 1;cur = ch;}}replace += cnt / 3;return Math.max(replace, 3 - categories);} else {// 替换次数和删除次数int replace = 0, remove = n - 20;// k mod 3 = 1 的组数，即删除 2 个字符可以减少 1 次替换操作int rm2 = 0;int cnt = 0;char cur = '#';for (int i = 0; i < n; ++i) {char ch = password.charAt(i);if (ch == cur) {++cnt;} else {if (remove > 0 && cnt >= 3) {if (cnt % 3 == 0) {// 如果是 k % 3 = 0 的组，那么优先删除 1 个字符，减少 1 次替换操作--remove;--replace;} else if (cnt % 3 == 1) {// 如果是 k % 3 = 1 的组，那么存下来备用++rm2;}// k % 3 = 2 的组无需显式考虑}replace += cnt / 3;cnt = 1;cur = ch;}}if (remove > 0 && cnt >= 3) {if (cnt % 3 == 0) {--remove;--replace;} else if (cnt % 3 == 1) {++rm2;}}replace += cnt / 3;// 使用 k % 3 = 1 的组的数量，由剩余的替换次数、组数和剩余的删除次数共同决定int use2 = Math.min(Math.min(replace, rm2), remove / 2);replace -= use2;remove -= use2 * 2;// 由于每有一次替换次数就一定有 3 个连续相同的字符（k / 3 决定），因此这里可以直接计算出使用 k % 3 = 2 的组的数量int use3 = Math.min(replace, remove / 3);replace -= use3;remove -= use3 * 3;return (n - 20) + Math.max(replace, 3 - categories);}}
}