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Refactor Code Style (#4151)
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@ -50,8 +50,8 @@ public class Anagrams {
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if (s.length() != t.length()) {
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return false;
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} else {
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char c[] = s.toCharArray();
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char d[] = t.toCharArray();
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char[] c = s.toCharArray();
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char[] d = t.toCharArray();
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Arrays.sort(c);
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Arrays.sort(
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d
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@ -65,8 +65,8 @@ public class Anagrams {
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if (a.length() != b.length()) {
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return false;
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} else {
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int m[] = new int[26];
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int n[] = new int[26];
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int[] m = new int[26];
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int[] n = new int[26];
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for (char c : a.toCharArray()) {
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m[c - 'a']++;
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}
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@ -90,8 +90,8 @@ public class Anagrams {
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}
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// this is similar to approach number 2 but here the string is not converted to character array
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else {
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int a[] = new int[26];
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int b[] = new int[26];
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int[] a = new int[26];
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int[] b = new int[26];
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int k = s.length();
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for (int i = 0; i < k; i++) {
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a[s.charAt(i) - 'a']++;
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@ -44,7 +44,7 @@ public class LetterCombinationsOfPhoneNumber {
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// Driver code
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public static void main(String[] args) {
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int number[] = { 2, 3, 4 };
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int[] number = { 2, 3, 4 };
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printWords(number);
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}
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}
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@ -22,18 +22,8 @@ public static int myAtoi(String s) {
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number = "0";
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break;
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}
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switch (ch) {
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case '0' -> number += ch;
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case '1' -> number += ch;
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case '2' -> number += ch;
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case '3' -> number += ch;
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case '4' -> number += ch;
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case '5' -> number += ch;
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case '6' -> number += ch;
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case '7' -> number += ch;
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case '8' -> number += ch;
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case '9' -> number += ch;
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}
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if(ch >= '0' && ch <= '9')
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number += ch;
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} else if (ch == '-' && !isDigit) {
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number += "0";
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negative = true;
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@ -48,10 +48,7 @@ class WordLadder {
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* if the endword is there. Otherwise, will return the length as 0.
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*/
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public static int ladderLength(String beginWord, String endWord, List<String> wordList) {
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HashSet<String> set = new HashSet();
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for (String word : wordList) {
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set.add(word);
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}
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HashSet<String> set = new HashSet(wordList);
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if (!set.contains(endWord)) {
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return 0;
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