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dede小说网站模板下载,wordpress调出代码,自助建站系统破解版,wordpress图片延迟加载题目描述 给你一个字符串str和整数k#xff0c;返回满足以下条件的所有子字符串个数#xff1a; 恰好包含k个字母。数字0-9各出现至少一次。 输入描述 第一行字符串str(1≤ length ≤ 100000),仅包含数字和小写字母第二行为整数k(0 ≤ k ≤100000 ) 输出描述 输出一个整…题目描述给你一个字符串str和整数k返回满足以下条件的所有子字符串个数恰好包含k个字母。数字0-9各出现至少一次。输入描述第一行字符串str(1≤ length ≤ 100000),仅包含数字和小写字母第二行为整数k(0 ≤ k ≤100000 )输出描述输出一个整数表示满足所有条件的子字符串的个数。子字符串是字符串中连续的非空字符序列示例1输入a0123456789aa 1输出2问题分析题目要求统计满足以下条件的子字符串个数子字符串恰好包含k个字母大小写不敏感题目中均为小写字母。子字符串中数字0-9各出现至少一次。解决思路滑动窗口使用滑动窗口技术高效遍历所有可能的子字符串。字母计数维护窗口内字母的数量确保恰好为k。数字覆盖确保窗口内包含所有数字0-9至少一次。优化检查在满足字母数量时检查数字覆盖情况。算法步骤初始化左右指针left和right表示窗口的左右边界。维护两个哈希表或数组letterCount统计字母数量digitCount统计数字出现情况。移动右指针扩展窗口更新计数。当字母数量等于k时检查数字是否覆盖0-9。移动左指针收缩窗口确保字母数量不超过k。代码实现C 实现#include iostream #include string #include unordered_map using namespace std; int countSubstrings(const string str, int k) { int n str.size(); int res 0; int left 0; unordered_mapchar, int letterCount; unordered_mapchar, int digitCount; int requiredDigits 10; for (int right 0; right n; right) { char c str[right]; if (isdigit(c)) { digitCount[c]; } else { letterCount[c]; } while (letterCount.size() k || (letterCount.size() k digitCount.size() requiredDigits)) { char leftChar str[left]; if (isdigit(leftChar)) { digitCount[leftChar]--; if (digitCount[leftChar] 0) { digitCount.erase(leftChar); } } else { letterCount[leftChar]--; if (letterCount[leftChar] 0) { letterCount.erase(leftChar); } } left; } if (letterCount.size() k digitCount.size() requiredDigits) { res; } } return res; } int main() { string str; int k; cin str k; cout countSubstrings(str, k) endl; return 0; }Python 实现def count_substrings(s, k): n len(s) res 0 left 0 letter_count {} digit_count {} required_digits 10 for right in range(n): c s[right] if c.isdigit(): digit_count[c] digit_count.get(c, 0) 1 else: letter_count[c] letter_count.get(c, 0) 1 while len(letter_count) k or (len(letter_count) k and len(digit_count) required_digits): left_char s[left] if left_char.isdigit(): digit_count[left_char] - 1 if digit_count[left_char] 0: digit_count.pop(left_char) else: letter_count[left_char] - 1 if letter_count[left_char] 0: letter_count.pop(left_char) left 1 if len(letter_count) k and len(digit_count) required_digits: res 1 return res s input().strip() k int(input()) print(count_substrings(s, k))Java 实现import java.util.HashMap; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner scanner new Scanner(System.in); String str scanner.nextLine(); int k scanner.nextInt(); System.out.println(countSubstrings(str, k)); } public static int countSubstrings(String str, int k) { int n str.length(); int res 0; int left 0; HashMapCharacter, Integer letterCount new HashMap(); HashMapCharacter, Integer digitCount new HashMap(); int requiredDigits 10; for (int right 0; right n; right) { char c str.charAt(right); if (Character.isDigit(c)) { digitCount.put(c, digitCount.getOrDefault(c, 0) 1); } else { letterCount.put(c, letterCount.getOrDefault(c, 0) 1); } while (letterCount.size() k || (letterCount.size() k digitCount.size() requiredDigits)) { char leftChar str.charAt(left); if (Character.isDigit(leftChar)) { digitCount.put(leftChar, digitCount.get(leftChar) - 1); if (digitCount.get(leftChar) 0) { digitCount.remove(leftChar); } } else { letterCount.put(leftChar, letterCount.get(leftChar) - 1); if (letterCount.get(leftChar) 0) { letterCount.remove(leftChar); } } left; } if (letterCount.size() k digitCount.size() requiredDigits) { res; } } return res; } }JavaScript 实现function countSubstrings(str, k) { let n str.length; let res 0; let left 0; let letterCount new Map(); let digitCount new Map(); const requiredDigits 10; for (let right 0; right n; right) { const c str[right]; if (/\d/.test(c)) { digitCount.set(c, (digitCount.get(c) || 0) 1); } else { letterCount.set(c, (letterCount.get(c) || 0) 1); } while (letterCount.size k || (letterCount.size k digitCount.size requiredDigits)) { const leftChar str[left]; if (/\d/.test(leftChar)) { digitCount.set(leftChar, digitCount.get(leftChar) - 1); if (digitCount.get(leftChar) 0) { digitCount.delete(leftChar); } } else { letterCount.set(leftChar, letterCount.get(leftChar) - 1); if (letterCount.get(leftChar) 0) { letterCount.delete(leftChar); } } left; } if (letterCount.size k digitCount.size requiredDigits) { res; } } return res; } const readline require(readline); const rl readline.createInterface({ input: process.stdin, output: process.stdout }); let input []; rl.on(line, (line) { input.push(line); if (input.length 2) { const str input[0]; const k parseInt(input[1]); console.log(countSubstrings(str, k)); rl.close(); } });C 实现#include stdio.h #include string.h #include ctype.h #define MAX_CHAR 128 int countSubstrings(const char *str, int k) { int n strlen(str); int res 0; int left 0; int letterCount[MAX_CHAR] {0}; int digitCount[MAX_CHAR] {0}; int uniqueLetters 0; int uniqueDigits 0; int requiredDigits 10; for (int right 0; right n; right) { char c str[right]; if (isdigit(c)) { if (digitCount[c] 0) { uniqueDigits; } digitCount[c]; } else { if (letterCount[c] 0) { uniqueLetters; } letterCount[c]; } while (uniqueLetters k || (uniqueLetters k uniqueDigits requiredDigits)) { char leftChar str[left]; if (isdigit(leftChar)) { digitCount[leftChar]--; if (digitCount[leftChar] 0) { uniqueDigits--; } } else { letterCount[leftChar]--; if (letterCount[leftChar] 0) { uniqueLetters--; } } left; } if (uniqueLetters k uniqueDigits requiredDigits) { res; } } return res; } int main() { char str[100001]; int k; scanf(%s %d, str, k); printf(%d\n, countSubstrings(str, k)); return 0; }代码说明滑动窗口通过左右指针动态调整窗口大小。哈希表/数组用于统计字母和数字的出现次数。条件检查确保字母数量为k且数字覆盖0-9。时间复杂度O(n)每个字符最多被访问两次左右指针各一次。空间复杂度O(1)使用固定大小的数组或哈希表。