// Prints all large anagram groups in a dictionary iteratively public class Anagrams { public static void main(String[] args) throws IOException { File dictionary = new File(args[0]); int minGroupSize = Integer.
public boolean isAnagram2(Strings,Stringt) {if(s ==null|| t ==null|| s.length() != t.length()) {returnfalse; }int[] arr =newint[26];for(inti=0; i<s.length(); i++) { arr[s.charAt(i)-'a']++; arr[t.charAt(i)-'a']--; }for(intnum: arr) {if(num!=0) {return...
For example, the word bad causes an entry mapping abd into bad to be put into the multimap. A moment's reflection will show that all the words to which any given key maps form an anagram group. It's a simple matter to iterate over the keys in the multimap, printing out each anagram...
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javaCopy codepublic boolean isAnagram(String s, String t) { if (s.length() != t.length()) { return false; } int[] counter = new int[26]; // 假设只包含小写字母 for (int i = 0; i < s.length(); i++) { counter[s.charAt(i) - 'a']++; ...
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https://leetcode.cn/problems/valid-anagram/submissions/ 【思路】 因为单词是由字母组成的,可以直接通过ASCII将字母看成是数字 public boolean isAnagram(String s, String t) {// 因为参数里面都是小写字母,因此只需要创建长度为26的数组即可int[] arr = new int[26];// 统计字符串1的每个字母的个数for...
public boolean isAnagram(String s, String t) { if (s.length() != t.length()) return false; int[] letterCount = new int[26]; //统计字符串s中的每个字符的数量 for (int i = 0; i < s.length(); i++) letterCount[s.charAt(i) - 'a']++; ...
程序应该忽略空白和标点符号 public class Anagram { public static boolean areAnagrams(String string1, String string2) { String workingCopy1 = removeJunk(string1); String workingCopy2 = removeJunk(string2); workingCopy1 = workingCopy1.toLowerCase(); workingCopy2 = workingCopy2.toLowerCase(); ...
// Prints all large anagram groups in a dictionary iteratively public class Anagrams { public static void main(String[] args) throws IOException { File dictionary = new File(args[0]); int minGroupSize = Integer.parseInt(args[1]); Map<String, Set<String>> groups = new HashMap<>(); try...