给定字符串s，我们必须找到s的最长子字符串的长度，该子字符串恰好包含K个不同的元音。
注意：将大写和小写字符视为两个不同的字符。
例子：
Input : s = “tHeracEBetwEEntheTwo”, k = 1
Output : 14
Explanation : Longest substring with only 1 vowel is “cEBetwEEntheTw”
and its length is 14.
Input : s = “artyebui”, k = 2
Output : 6
Explanation : Longest substring with only 2 vowel is “rtyebu”
蛮力法：对于每个子串，我们检查K个不同元音的条件，并检查其长度。最后，最大长度将是结果。
高效的方法：在这里，我们保持子串中出现的元音计数。直到K不为零，我们才算出子串中出现的不同元音。当K变为负数时，我们开始删除直到那个时候为止发现的子串的第一个元音，以便之后可能再有新的子串(更大的length)成为可能。删除元音时，我们会减少其数量，以便新的子字符串可以包含出现在字符串后半部分的元音。当K为0时，我们得到子串的长度。
下面是上述方法的实现
C++
// CPP program to find the longest substring
// with k distinct vowels.
#include
using namespace std;
#define MAX 128
// Function to check whether a character is
// vowel or not
bool isVowel(char x)
{
return (x == 'a' || x == 'e' || x == 'i' ||
x == 'o' || x == 'u' || x == 'A' ||
x == 'E' || x == 'I' || x == 'O' ||
x == 'U');
}
int KDistinctVowel(char s[], int k)
{
// length of string
int n = strlen(s);
// array for count of characters
int c[MAX];
memset(c, 0, sizeof(c));
// Initialize result to be
// negative
int result = -1;
for (int i = 0, j = -1; i < n; ++i) {
int x = s[i];
// If letter is vowel then we
// increment its count value
// and decrease the k value so
// that if we again encounter the
// same vowel character then we
// don't consider it for our result
if (isVowel(x)) {
if (++c[x] == 1) {
// Decrementing the K value
--k;
}
}
// Till k is 0 above if condition run
// after that this while loop condition
// also become active. Here what we have
// done actually is that, if K is less
// than 0 then we eliminate the first
// vowel we have encountered till that
// time . Now K is incremented and k
// becomes 0. So, now we calculate the
// length of substring from the present
// index to the deleted index of vowel
// which result in our results.
while (k < 0) {
x = s[++j];
if (isVowel(x)) {
// decresing the count
// so that it may appear
// in another substring
// appearing after this
// present substring
if (--c[x] == 0) {
// incrementing the K value
++k;
}
}
}
// Checking the maximum value
// of the result by comparing
// the length of substring
// whenever K value is 0 means
// K distinct vowel is present
// in substring
if (k == 0)
result = max(result, i - j);
}
return result;
}
// Driver code
int main(void)
{
char s[] = "tHeracEBetwEEntheTwo";
int k = 1;
cout << KDistinctVowel(s, k);
return 0;
} Java
// Java program to find the longest substring
// with k distinct vowels.
class GFG {
static int MAX = 128;
// Function to check whether a character is
// vowel or not
static boolean isVowel(char x) {
return (x == 'a' || x == 'e' || x == 'i' ||
x == 'o' || x == 'u' || x == 'A' ||
x == 'E' || x == 'I' || x == 'O' ||
x == 'U');
}
static int KDistinctVowel(String s, int k) {
// length of string
int n = s.length();
// array for count of characters
int[] c = new int[MAX];
//Array.Clear(c, 0, c.Length);
// Initialize result to be
// negative
int result = -1;
for (int i = 0, j = -1; i < n; ++i) {
char x = s.charAt(i);
// If letter is vowel then we
// increment its count value
// and decrease the k value so
// that if we again encounter the
// same vowel character then we
// don't consider it for our result
if (isVowel(x)) {
if (++c[x] == 1) {
// Decrementing the K value
--k;
}
}
// Till k is 0 above if condition run
// after that this while loop condition
// also become active. Here what we have
// done actually is that, if K is less
// than 0 then we eliminate the first
// vowel we have encountered till that
// time . Now K is incremented and k
// becomes 0. So, now we calculate the
// length of substring from the present
// index to the deleted index of vowel
// which result in our results.
while (k < 0) {
x = s.charAt(++j);
if (isVowel(x)) {
// decresing the count
// so that it may appear
// in another substring
// appearing after this
// present substring
if (--c[x] == 0) {
// incrementing the K value
++k;
}
}
}
// Checking the maximum value
// of the result by comparing
// the length of substring
// whenever K value is 0 means
// K distinct vowel is present
// in substring
if (k == 0) {
result = Math.max(result, i - j);
}
}
return result;
}
// Driver code
public static void main(String[] args) {
String s = "tHeracEBetwEEntheTwo";
int k = 1;
System.out.println(KDistinctVowel(s, k));
}
}
/* This Java code is contributed by Rajput-Ji*/Python3
# Python3 program to find the longest substring
# with k distinct vowels.
MAX = 128
# Function to check whether a character is
# vowel or not
def isVowel(x):
return (x == 'a' or x == 'e' or x == 'i' or
x == 'o' or x == 'u' or x == 'A' or
x == 'E' or x == 'I' or x == 'O' or
x == 'U')
def KDistinctVowel(c,k):
n = len(s)
c = [0 for i in range(MAX)]
result = -1
j = -1
for i in range(n):
x=s[i]
# If letter is vowel then we
# increment its count value
# and decrease the k value so
# that if we again encounter the
# same vowel character then we
# don't consider it for our result
if isVowel(x):
c[ord(x)] += 1
if c[ord(x)] == 1:
k -= 1
# Till k is 0 above if condition run
# after that this while loop condition
# also become active. Here what we have
# done actually is that, if K is less
# than 0 then we eliminate the first
# vowel we have encountered till that
# time . Now K is incremented and k
# becomes 0. So, now we calculate the
# length of substring from the present
# index to the deleted index of vowel
# which result in our results.
while k < 0:
j += 1
x = s[j]
if isVowel(x):
# decresing the count
# so that it may appear
# in another substring
# appearing after this
# present substring
c[ord(x)] -= 1
k += 1
# Checking the maximum value
# of the result by comparing
# the length of substring
# whenever K value is 0 means
# K distinct vowel is present
# in substring
if k == 0:
result = max(result, i - j)
return result
s = "tHeracEBetwEEntheTwo"
k = 1
print(KDistinctVowel(s, k))
# This code is contributed by mohit kumar 29C#
// C# program to find the longest substring
// with k distinct vowels.
using System;
class GFG {
static int MAX = 128;
// Function to check whether a character is
// vowel or not
static bool isVowel(char x)
{
return (x == 'a' || x == 'e' || x == 'i' ||
x == 'o' || x == 'u' || x == 'A' ||
x == 'E' || x == 'I' || x == 'O' ||
x == 'U');
}
static int KDistinctVowel(string s, int k)
{
// length of string
int n = s.Length;
// array for count of characters
int []c = new int[MAX];
Array.Clear(c, 0, c.Length);
// Initialize result to be
// negative
int result = -1;
for (int i = 0, j = -1; i < n; ++i) {
char x = s[i];
// If letter is vowel then we
// increment its count value
// and decrease the k value so
// that if we again encounter the
// same vowel character then we
// don't consider it for our result
if (isVowel(x)) {
if (++c[x] == 1) {
// Decrementing the K value
--k;
}
}
// Till k is 0 above if condition run
// after that this while loop condition
// also become active. Here what we have
// done actually is that, if K is less
// than 0 then we eliminate the first
// vowel we have encountered till that
// time . Now K is incremented and k
// becomes 0. So, now we calculate the
// length of substring from the present
// index to the deleted index of vowel
// which result in our results.
while (k < 0) {
x = s[++j];
if (isVowel(x)) {
// decresing the count
// so that it may appear
// in another substring
// appearing after this
// present substring
if (--c[x] == 0) {
// incrementing the K value
++k;
}
}
}
// Checking the maximum value
// of the result by comparing
// the length of substring
// whenever K value is 0 means
// K distinct vowel is present
// in substring
if (k == 0) {
result = Math.Max(result, i - j);
}
}
return result;
}
// Driver code
static void Main()
{
string s = "tHeracEBetwEEntheTwo";
int k = 1;
Console.Write(KDistinctVowel(s, k));
}
}
// This code is contributed Manish Shaw
// (manishshaw1)PHP
输出：
7