给定一个人的名字和键入的名字。有时，在输入元音[aeiou]时，按键可能会被长按，并且字符会被键入1次或多次。任务是检查键入的名称，并判断键入的名称是否可能与人的名字相同，并且长按了某些字符(可能没有字符)。如果是，则返回“ True ”，否则返回“ False ”。
注意： name和typed-name用空格分隔，个人名称之间没有空格。名称的每个字符都是唯一的。
例子：
Input: str = “geeks”, typed = “geeeeks”
Output: True
The vowel ‘e’ repeats more times in typed and all other characters match.
Input: str = “alice”, typed = “aallicce”
Output: False
Here ‘l’ and ‘c’ are repeated which not a vowel.
Hence name and typed-name represents different names.
Input: str = “alex”, typed = “aaalaeex”
Output: False
A vowel ‘a’ is extra in typed.
方法：该思想基于游程长度编码。我们只考虑元音，并计算它们在str和type中的连续出现次数。 str中出现的次数必须更少。
下面是上述方法的实现。
C++
// CPP program to implement run length encoding
#include
using namespace std;
// Check if the character is vowel or not
bool isVowel(char c)
{
string vowel = "aeiou";
for (int i = 0; i < vowel.length(); ++i)
if (vowel[i] == c)
return true;
return false;
}
// Returns true if 'typed' is a typed name
// given str
bool printRLE(string str, string typed)
{
int n = str.length(), m = typed.length();
// Traverse through all characters of str.
int j = 0;
for (int i = 0; i < n; i++) {
// If current characters do not match
if (str[i] != typed[j])
return false;
// If not vowel, simply move ahead in both
if (isVowel(str[i]) == false) {
j++;
continue;
}
// Count occurrences of current vowel in str
int count1 = 1;
while (i < n - 1 && str[i] == str[i + 1]) {
count1++;
i++;
}
// Count occurrences of current vowel in
// typed.
int count2 = 1;
while (j < m - 1 && typed[j] == str[i]) {
count2++;
j++;
}
if (count1 > count2)
return false;
}
return true;
}
int main()
{
string name = "alex", typed = "aaalaeex";
if (printRLE(name, typed))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java program to implement run length encoding
public class Improve {
// Check if the character is vowel or not
static boolean isVowel(char c)
{
String vowel = "aeiou";
for (int i = 0; i < vowel.length(); ++i)
if (vowel.charAt(i) == c)
return true;
return false;
}
// Returns true if 'typed' is a typed name
// given str
static boolean printRLE(String str, String typed)
{
int n = str.length(), m = typed.length();
// Traverse through all characters of str.
int j = 0;
for (int i = 0; i < n; i++) {
// If current characters do not match
if (str.charAt(i) != typed.charAt(j))
return false;
// If not vowel, simply move ahead in both
if (isVowel(str.charAt(i)) == false) {
j++;
continue;
}
// Count occurrences of current vowel in str
int count1 = 1;
while (i < n - 1 && str.charAt(i) == str.charAt(i+1)) {
count1++;
i++;
}
// Count occurrences of current vowel in
// typed.
int count2 = 1;
while (j < m - 1 && typed.charAt(j) == str.charAt(i)) {
count2++;
j++;
}
if (count1 > count2)
return false;
}
return true;
}
public static void main(String args[])
{
String name = "alex", typed = "aaalaeex";
if (printRLE(name, typed))
System.out.println("Yes");
else
System.out.println("No");
}
// This code is contributed by ANKITRAI1
}Python3
# Python3 program to implement run
# length encoding
# Check if the character is
# vowel or not
def isVowel(c):
vowel = "aeiou"
for i in range(len(vowel)):
if(vowel[i] == c):
return True
return False
# Returns true if 'typed' is a
# typed name given str
def printRLE(str, typed):
n = len(str)
m = len(typed)
# Traverse through all
# characters of str
j = 0
for i in range(n):
# If current characters do
# not match
if str[i] != typed[j]:
return False
# If not vowel, simply move
# ahead in both
if isVowel(str[i]) == False:
j = j + 1
continue
# Count occurences of current
# vowel in str
count1 = 1
while (i < n - 1 and (str[i] == str[i + 1])):
count1 = count1 + 1
i = i + 1
# Count occurence of current
# vowel in typed
count2 = 1
while(j < m - 1 and typed[j] == str[i]):
count2 = count2 + 1
j = j + 1
if count1 > count2:
return False
return True
# Driver code
name = "alex"
typed = "aaalaeex"
if (printRLE(name, typed)):
print("Yes")
else:
print("No")
# This code is contributed
# by Shashank_SharmaC#
// C# program to implement run
// length encoding
using System;
class GFG
{
// Check if the character is
// vowel or not
public static bool isVowel(char c)
{
string vowel = "aeiou";
for (int i = 0;
i < vowel.Length; ++i)
{
if (vowel[i] == c)
{
return true;
}
}
return false;
}
// Returns true if 'typed' is
// a typed name given str
public static bool printRLE(string str,
string typed)
{
int n = str.Length, m = typed.Length;
// Traverse through all
// characters of str.
int j = 0;
for (int i = 0; i < n; i++)
{
// If current characters
// do not match
if (str[i] != typed[j])
{
return false;
}
// If not vowel, simply move
// ahead in both
if (isVowel(str[i]) == false)
{
j++;
continue;
}
// Count occurrences of current
// vowel in str
int count1 = 1;
while (i < n - 1 &&
str[i] == str[i + 1])
{
count1++;
i++;
}
// Count occurrences of current
// vowel in typed
int count2 = 1;
while (j < m - 1 &&
typed[j] == str[i])
{
count2++;
j++;
}
if (count1 > count2)
{
return false;
}
}
return true;
}
// Driver Code
public static void Main(string[] args)
{
string name = "alex",
typed = "aaalaeex";
if (printRLE(name, typed))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed
// by Shrikant13PHP
$count2)
return false;
}
return true;
}
// Driver code
$name = "alex";
$typed = "aaalaeex";
if (printRLE($name, $typed))
echo "Yes";
else
echo "No";
// This code is contributed
// by Shivi_Aggarwal
?>输出：
No
时间复杂度： O(m + n)
辅助空间： O(1)