LeetCode笔记:438. Find All Anagrams in a String 我自己的做法在超长的测试用例时超时了,用的循环太多了。这里看别人非常精简巧妙的一个方法。...,这里也巧妙地利用了这个特性。...这个代码可能过于精简了,来看一下稍微复杂化一点的同样的代码。...s.charAt(left)]++; left++; } } return
import java.util.ArrayList; import java.util.List; public class FindAllAnagrams { public List<Integer> findAnagrams(String s, String p) { List<Integer> result = new ArrayList<>(); if (s.length() < p.length()) return result; int[] p_count = new int[26]; int[] s_count = new ...
意思是看两个string中各个字母出现次数是否相同,用数组存储就行(哈希),结果: Success Runtime: 8 ms, faster than 98.42% of C++ online submissions for Valid Anagram. Memory Usage: 9.4 MB, less than 50.10% of C++ online submissions for Valid Anagram. 代码: class Solution { public: bool isAnagram...
In unserem Beispiel unten veranschaulichen wir, wie wir Anagramme in Java mit bitweisemXORfinden können. Der Code wird wie folgt sein: publicclassJavaAnagram{publicstaticvoidmain(String[]args){// Declaring two stringString STR_1="Race";String STR_2="Care";if(AnagramChecking(STR_1,STR_2...
public bool IsAnagram2(string s, string t) { //本题都是小写 //利用ASCII码 //每个字母一个坑 if (s.Length != t.Length) { return false; } int[] ss = new int[26]; foreach (char schar in s) { ss[schar - 'a']++; } foreach (char tchar in t) { ss[tchar - 'a']--;...
publicbooleanisAnagram(Strings,Stringt) {if(s ==null|| t ==null|| s.length() != t.length()) {returnfalse; } char[] ch = s.toCharArray(); char[] ch2 = t.toCharArray();Arrays.sort(ch);Arrays.sort(ch2);returnArrays.equals(ch, ch2); ...
Leetcode.242 Valid AnagramGiven two strings s and t , write a function to determine if t is an anagram of s.Example 1:Input: s = "anagram", t = "nagaram" Output: true Example 2:Input: s = "rat", t = "car" Output: false Note: You may assume the string contains only ...
Then store the count as value in the integer array Iterate and check for non zero count **/ class Solution { public boolean isAnagram(String s, String t) { Map<Character, Integer> charCountMap = new HashMap<>(); for(char c: s.toCharArray()) { charCountMap.put(c, charCountMap.get...
//{{{ #include #include <algorithm> #include <iostream> #include <cstring> #include <vector> #include <cstdio> #include <string> #include <cmath> #include <queue> #include <set> #include <map> #include <complex> //}}}//#include <bits/stdc++.h> using namesp ...
Once the strings are in lowercase, we sort the characters in each string and then join them back into new strings. This transformation ensures that if the strings are anagrams, they will now be identical. cpp csharp go java javascript python 1a = a.split('').sort().join(''); 2b = ...