leetcode算法之Valid Anagram

原文算法说明如下：

Given two strings s and t, write a function to determine if t is an anagram of s.

For example,s = "anagram", t = "nagaram", return true.s = "rat", t = "car", return false.

翻译：判断给定的两个字符串是否为打乱了顺序的同一个字符串

我的java实现：

import java.util.HashMap; public class Solution { public boolean isAnagram(String s, String t) { if(s.length() != t.length()){ return false; } char[] c1 = s.toCharArray(); char[] c2 = t.toCharArray(); HashMap<Character,Integer> map1 = new HashMap<Character,Integer>(); HashMap<Character,Integer> map2 = new HashMap<Character,Integer>(); for(int i=0;i < s.length();i++){ if(!map1.containsKey(c1[i])){ map1.put(c1[i], 1); }else{ map1.put(c1[i], map1.get(c1[i])+1); } if(!map2.containsKey(c2[i])){ map2.put(c2[i], 1); }else{ map2.put(c2[i], map2.get(c2[i])+1); } } if(map1.entrySet().containsAll(map2.entrySet()) && map2.entrySet().containsAll(map1.entrySet())){ return true; } return false; } }

转载于:https://www.cnblogs.com/iamconan/p/7383578.html

相关资源：JAVA上百实例源码以及开源项目