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法律研究检查字符串是否可以拆分为两个元音数量相等的子字符串
欢迎来到另一篇关于 C++ 中一个引人入胜的解决问题主题的深入指南。这次，我们将解决确定一个字符串是否可以分成两个子字符串的问题，每个子字符串包含相同数量的元音。这个问题是磨练您在字符串操作和元音计数方面的技能的绝佳练习。
问题陈述
给定一个字符串，我们的目标是确定它是否可以被划分为两个非空子字符串，使得这两个子字符串具有相同数量的元音。英语字母表中的元音是 'a'、'e'、'i'、'o'、'u'、'A'、'E'、'I'、'O'、'U'。
方法
我们的方法是首先计算字符串中元音的总数。如果总数不是偶数，我们立即知道无法将字符串拆分为两个元音数量相等的子字符串。
如果总数是偶数，那么我们遍历字符串，并对遇到的元音进行计数。如果在任何时候我们的计数等于总数的一半，我们可以在该点将字符串拆分为两个元音数量相等的子字符串。
示例
以下是上述方法的程序 -
#include <stdio.h>
#include <stdbool.h>
// Function to check if a character is a vowel.
bool isVowel(char c) {
return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');
}
// Function to determine if a string can be split into two substrings with an equal number of vowels.
bool canBeSplit(char *s) {
int totalVowels = 0;
// Count the total number of vowels in the string.
for (int i = 0; s[i] != '\0'; i++) {
if (isVowel(s[i]))
totalVowels++;
}
if (totalVowels % 2 != 0)
return false;
int halfVowels = 0;
for (int i = 0; s[i] != '\0'; i++) {
if (isVowel(s[i]))
halfVowels++;
if (halfVowels == totalVowels / 2)
return true; // If half the vowels are encountered, return true.
}
return false; // If half the vowels aren't encountered by the end, return false.
}
int main() {
char s[] = "CIVIC";
if (canBeSplit(s))
printf("Yes, the string can be split into two substrings with an equal number of vowels.\n");
else
printf("No, the string cannot be split into two substrings with an equal number of vowels.\n");
return 0;
}
输出
Yes, the string can be split into two substrings with an equal number of vowels.
#include<iostream>
using namespace std;
// Function to check if a character is a vowel.
bool isVowel(char c) {
return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');
}
// Function to determine if a string can be split into two substrings with an equal number of vowels.
bool canBeSplit(string s) {
int totalVowels = 0;
// Count the total number of vowels in the string.
for (char c : s) {
if (isVowel(c))
totalVowels++;
}
if (totalVowels % 2 != 0)
return false;
int halfVowels = 0;
for (char c : s) {
if (isVowel(c))
halfVowels++;
if (halfVowels == totalVowels / 2)
return true; // If half the vowels are encountered, return true.
}
return false; // If half the vowels aren't encountered by the end, return false.
}
int main() {
string s = "CIVIC";
if (canBeSplit(s))
cout << "Yes, the string can be split into two substrings with an equal number of vowels." << endl;
else
cout << "No, the string cannot be split into two substrings with an equal number of vowels." << endl;
return 0;
}
输出
Yes, the string can be split into two substrings with an equal number of vowels.
public class VowelSplit {
// Function to check if a character is a vowel.
static boolean isVowel(char c) {
return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');
}
// Function to determine if a string can be split into two substrings with an equal number of vowels.
static boolean canBeSplit(String s) {
int totalVowels = 0;
// Count the total number of vowels in the string.
for (char c : s.toCharArray()) {
if (isVowel(c))
totalVowels++;
}
if (totalVowels % 2 != 0)
return false;
int halfVowels = 0;
for (char c : s.toCharArray()) {
if (isVowel(c))
halfVowels++;
if (halfVowels == totalVowels / 2)
return true; // If half the vowels are encountered, return true.
}
return false; // If half the vowels aren't encountered by the end, return false.
}
public static void main(String[] args) {
String s = "CIVIC";
if (canBeSplit(s))
System.out.println("Yes, the string can be split into two substrings with an equal number of vowels.");
else
System.out.println("No, the string cannot be split into two substrings with an equal number of vowels.");
}
}
输出
Yes, the string can be split into two substrings with an equal number of vowels.
def is_vowel(c):
return c in 'aeiouAEIOU'
def can_be_split(s):
total_vowels = sum(1 for c in s if is_vowel(c))
if total_vowels % 2 != 0:
return False
half_vowels = 0
for c in s:
if is_vowel(c):
half_vowels += 1
if half_vowels == total_vowels / 2:
return True
return False
s = "CIVIC"
if can_be_split(s):
print("Yes, the string can be split into two substrings with an equal number of vowels.")
else:
print("No, the string cannot be split into two substrings with an equal number of vowels.")
输出
Yes, the string can be split into two substrings with an equal number of vowels.
测试用例示例
让我们用一个例子来说明这个问题及其解决方案 -
假设字符串为“beautiful”。
我们首先计算“beautiful”中元音的总数，即 5。由于这不是偶数，我们立即知道该字符串不能拆分为两个元音数量相等的子字符串。
因此，输出将是“否，该字符串不能拆分为两个元音数量相等的子字符串。”
结论
通过本 C++ 指南，我们学习了如何检查字符串是否可以分成两个子字符串，以便每个子字符串包含相同数量的元音。此问题是在 C++ 语言中进行字符串操作和字符计数的有用练习。
更新于: 2023年10月16日
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