如果字符串中的所有字母都相同，则尝试返回 true

Question

What I have so far:

 public boolean allSameLetter(String str)
{
  for (int i = 1; i < str.length(); i++)
    {
        int charb4 = i--;
        if ( str.charAt(i) != str.charAt(charb4))
        {
        return false;
        }

        if ( i == str.length())
        {
        return true;
        }
    } 
}

Please excuse any inefficiencies if any; still relatively new to coding in general. Am I lacking some knowledge in terms of using operators and .charAt() together? Is it illogical? Or is my error elsewhere?

Answer 1

Using regex:

return str.matches("^(.)\\1*$");

Using streams:

str.chars().allMatch(c -> c == str.charAt(0));

Other:

return str.replace(String.valueOf(str.charAt(0), "").length() == 0;

Answer 2

You can follow the below steps:

(1) Get the first character (ie, 0th index)

(2) Check the first character is the same with subsequent characters, if not return false (and comes out from method)

(3) If all chars match ie, processing goes till the end of the method and returns true

  public boolean allSameLetter(String str) {
  char c1 = str.charAt(0);
  for(int i=1;i<str.length;i++) {
      char temp = str.charAt(i);
      if(c1 != temp) {
         //if chars does NOT match, 
         //just return false from here itself,
         //there is no need to verify other chars
         return false;
      }
  }
  //As it did NOT return from above if (inside for)
  //it means, all chars matched, so return true
  return true;
}

Answer 3

As Andrew said, you are decreasing i within your for loop. You can fix this by changing it to int charb4 = i - 1; . As for making your code more efficient you could condense it down to this.

public boolean allSameLetter(String str) {
    for(char c : str.toCharArray())
        if(c != str.charAt(0)) return false;
    return true;
}

Answer 4

Comment if you don't understand a part of it :)

 public boolean allSameLetter(String str)
 {
 for (int i = 1; i < str.length() -1; i++)
  {
    if ( str.charAt(i) != str.charAt(i+1))
    {
    return false;
    }
  } 
 return true
}

-1 is there since I am checking the current value in the array, then the next value in the array, thus I need to stop a place earlier.

If the loop if statement is never entered, it will make it far enough into the code to return true

Answer 5

You have to create a for loop that searches through the length of the String - 1. this way the program will not crash because of a 3 letter word with the program trying to get the 4th letter. This is what works for me:

public boolean allSameLetter(String str)
{
    for(int i = 0; i< str.length()-1; i++){
        if (str.charAt(i) != str.charAt(i+1)){
            return false;
        }
    }
    return true;
}

Answer 6

if((new HashSet<Character>(Arrays.asList(s.toCharArray()))).size()==1) return true; return false;

这应该足够了

Answer 7

The bug is caused by

int charb4 = i--;

this line is equal to

int charb4 = i-1;
i=i-1;

Because of this, your loop will never stop.
The easiest way to fix this

public boolean allSameLetter(String str)
{
  for (int i = 1; i < str.length(); i++)
  {
     if ( str.charAt(i) != str.charAt(i-1))
     {
       return false;
     }
  } 
}