给定长度为N 的字符串str ，任务是使用二进制搜索技术找到仅包含元音的最长子字符串。
例子：
Input: str = “baeicba”
Output: 3
Explanation:
Longest substring which contains vowels only is “aei”.
Input: str = “aeiou”
Output: 5
方法：请参阅元音的最长子串以获取 O(N) 复杂度的方法。
二分搜索方法：在本文中，我们使用基于二分搜索的方法：
请按照以下步骤解决问题：
- 对范围从 1 到N的长度应用二分搜索。
- 对于每个mid值，检查是否存在长度为mid的子串，该子串中仅包含元音。
- 如果存在长度为mid的子串，则更新max的值并将l更新为mid+1以检查是否存在长度大于mid 且仅由元音组成的子串。
- 如果不存在这样的长度为mid的子串，则将r更新为mid-1以检查是否存在长度小于mid的仅由元音组成的子串。
- 重复以上三步，直到l小于等于r。
- 返回最后得到的最大长度。
下面是上述方法的实现：
C++
// C++ implementation of
// the above approach
#include
using namespace std;
// Function to check if a character
// is vowel or not
bool vowel(int vo)
{
// 0-a 1-b 2-c and so on 25-z
if (vo == 0 || vo == 4
|| vo == 8 || vo == 14
|| vo == 20)
return true;
else
return false;
}
// Function to check if any
// substring of length k exists
// which contains only vowels
bool check(string s, int k)
{
vector cnt(26, 0);
for (int i = 0; i < k - 1; i++) {
cnt[s[i] - 'a']++;
}
// Applying sliding window to get
// all substrings of length K
for (int i = k - 1; i < s.size();
i++) {
cnt[s[i] - 'a']++;
int flag1 = 0;
for (int j = 0; j < 26; j++) {
if (vowel(j) == false
&& cnt[j] > 0) {
flag1 = 1;
break;
}
}
if (flag1 == 0)
return true;
// Remove the occurence of
// (i-k+1)th character
cnt[s[i - k + 1] - 'a']--;
}
return false;
}
// Function to perform Binary Search
int longestSubstring(string s)
{
int l = 1, r = s.size();
int maxi = 0;
// Doing binary search on the lengths
while (l <= r) {
int mid = (l + r) / 2;
if (check(s, mid)) {
l = mid + 1;
maxi = max(maxi, mid);
}
else
r = mid - 1;
}
return maxi;
}
// Driver Code
int main()
{
string s = "sedrewaefhoiu";
cout << longestSubstring(s);
return 0;
} Java
// Java implementation of
// the above approach
import java.util.*;
class GFG{
// Function to check if a character
// is vowel or not
static boolean vowel(int vo)
{
// 0-a 1-b 2-c and so on 25-z
if (vo == 0 || vo == 4 ||
vo == 8 || vo == 14 ||
vo == 20)
return true;
else
return false;
}
// Function to check if any
// subString of length k exists
// which contains only vowels
static boolean check(String s, int k)
{
int []cnt = new int[26];
for (int i = 0; i < k - 1; i++)
{
cnt[s.charAt(i) - 'a']++;
}
// Applying sliding window to get
// all subStrings of length K
for (int i = k - 1; i < s.length(); i++)
{
cnt[s.charAt(i) - 'a']++;
int flag1 = 0;
for (int j = 0; j < 26; j++)
{
if (vowel(j) == false && cnt[j] > 0)
{
flag1 = 1;
break;
}
}
if (flag1 == 0)
return true;
// Remove the occurence of
// (i-k+1)th character
cnt[s.charAt(i - k + 1) - 'a']--;
}
return false;
}
// Function to perform Binary Search
static int longestSubString(String s)
{
int l = 1, r = s.length();
int maxi = 0;
// Doing binary search on the lengths
while (l <= r)
{
int mid = (l + r) / 2;
if (check(s, mid))
{
l = mid + 1;
maxi = Math.max(maxi, mid);
}
else
r = mid - 1;
}
return maxi;
}
// Driver Code
public static void main(String[] args)
{
String s = "sedrewaefhoiu";
System.out.print(longestSubString(s));
}
}
// This code is contributed by sapnasingh4991Python3
# Python3 implementation of
# the above approach
# Function to check if a character
# is vowel or not
def vowel(vo):
# 0-a 1-b 2-c and so on 25-z
if (vo == 0 or vo == 4 or
vo == 8 or vo == 14 or
vo == 20):
return True
else:
return False
# Function to check if any
# substring of length k exists
# which contains only vowels
def check(s, k):
cnt = [0] * 26
for i in range (k - 1):
cnt[ord(s[i]) - ord('a')] += 1
# Applying sliding window to get
# all substrings of length K
for i in range (k - 1, len(s)):
cnt[ord(s[i]) - ord('a')] += 1
flag1 = 0
for j in range (26):
if (vowel(j) == False
and cnt[j] > 0):
flag1 = 1
break
if (flag1 == 0):
return True
# Remove the occurence of
# (i-k+1)th character
cnt[ord(s[i - k + 1]) - ord('a')] -= 1
return False
# Function to perform Binary Search
def longestSubstring(s):
l = 1
r = len(s)
maxi = 0
# Doing binary search on the lengths
while (l <= r):
mid = (l + r) // 2
if (check(s, mid)):
l = mid + 1
maxi = max(maxi, mid)
else:
r = mid - 1
return maxi
# Driver Code
if __name__ == "__main__":
s = "sedrewaefhoiu"
print (longestSubstring(s))
# This code is contributed by ChitranayalC#
// C# implementation of
// the above approach
using System;
class GFG{
// Function to check if a character
// is vowel or not
static bool vowel(int vo)
{
// 0-a 1-b 2-c and so on 25-z
if (vo == 0 || vo == 4 ||
vo == 8 || vo == 14 ||
vo == 20)
return true;
else
return false;
}
// Function to check if any
// subString of length k exists
// which contains only vowels
static bool check(String s, int k)
{
int []cnt = new int[26];
for (int i = 0; i < k - 1; i++)
{
cnt[s[i] - 'a']++;
}
// Applying sliding window to get
// all subStrings of length K
for (int i = k - 1; i < s.Length; i++)
{
cnt[s[i] - 'a']++;
int flag1 = 0;
for (int j = 0; j < 26; j++)
{
if (vowel(j) == false && cnt[j] > 0)
{
flag1 = 1;
break;
}
}
if (flag1 == 0)
return true;
// Remove the occurence of
// (i-k+1)th character
cnt[s[i - k + 1] - 'a']--;
}
return false;
}
// Function to perform Binary Search
static int longestSubString(String s)
{
int l = 1, r = s.Length;
int maxi = 0;
// Doing binary search on the lengths
while (l <= r)
{
int mid = (l + r) / 2;
if (check(s, mid))
{
l = mid + 1;
maxi = Math.Max(maxi, mid);
}
else
r = mid - 1;
}
return maxi;
}
// Driver Code
public static void Main(String[] args)
{
String s = "sedrewaefhoiu";
Console.Write(longestSubString(s));
}
}
// This code is contributed by sapnasingh4991输出：
3
时间复杂度： O(NlogN)
辅助空间： O(N)
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