Example of Golang string anagram in detail

  • 2020-06-07 04:39:07
  • OfStack

Achieve the goal

The goal of this article is to write a function anagram(s, t) to determine if two strings are formed alphabetically. Without further ado, let's take a look at the details.

GoLang implementation


func solution(s , t string)bool{
 if s == t {
 return true
 }
 length := len(s)
 if length != len(t) {
 return false
 }
 //' ' 32 --> ~ 126
 const MAX_ASCII int= 94
 const SPACE_INDEX rune = 32

 numbers := [MAX_ASCII]int{}
 sRune := []rune(s)
 tRune :=[]rune(t)

 for i := 0 ; i < length ; i++ {
 index := tRune[i] - SPACE_INDEX
 numbers[index]++

 index = sRune[i] - SPACE_INDEX
 numbers[index]--
 }


 for i := 0 ; i < MAX_ASCII / 2 ; i++{
 mergeSize := numbers[i]
 if mergeSize != 0 || mergeSize != numbers[MAX_ASCII - 1 - i]{
  return false
 }
 }
 return true
}

Key point 1:

Defines the value that holds the last value to determine whether two strings are the same length:

According to the ASCII table:

The difference between the first single character "" in decimal 32 and the last single character" ~ "in decimal 126 is 94.

The prediction here is that each character has been used, so the length is simply defined as 94.

The Java implementation is similar to the above:


public boolean anagram(String s, String t) {
 if (s == null || t == null || s.length() ==0 || s.length() != t.length()){
  return false;
 }
 if (s.equals(t))return true;

 final int MAX_ASCII = 94;
 final char SPACE_INDEX = ' ';

 int[] numbers = new int[MAX_ASCII];
 int length = s.length();

 char[] sCharArray = s.toCharArray();
 char[] tCharArray = t.toCharArray();

 for(int i = 0 ; i< length ; i++){
  int index = sCharArray[i] - SPACE_INDEX;
  numbers[index]++;

  index = tCharArray[i] - SPACE_INDEX;
  numbers[index]--;
 }

 for (int i =0 ; i < MAX_ASCII / 2 ; i++ ) {
  int mergeSize = numbers[i];
  if ( mergeSize != 0 || mergeSize != numbers[MAX_ASCII - 1 - i]){
  return false;
  }
 }
 return true;
 }

conclusion


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