LeetCode 242. Valid Anagram

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题目

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.

Note:
You may assume the string contains only lowercase alphabets.

import java.util.HashMap;/** * Created by caowei on 16/5/18. */public class Solution242 {    public static void main(String args[]){        Solution242 solution242 = new Solution242();        String string1 = "anagram";        String string2 = "nagaram";        boolean isAnagram = solution242.isAnagram(string1, string2);        System.out.print(isAnagram);    }    /**     *我是用hashmap保存第一个字符串里面每一个字符对应的出现次数,然后遍历第二个字符串的每一个字符,     * 如果字符在hashmap中出现,则hashmap对应字符减一.     */    public boolean isAnagram(String s, String t) {        boolean isAnagram = true;        char char_s[] = s.toCharArray();        char char_t[] = t.toCharArray();        if(char_s.length!=char_t.length){            return  false;        }        HashMap<Character, Integer> hashMap_s = new HashMap<Character, Integer>();        for(int i=0;i<char_s.length;i++){            if(hashMap_s.get(char_s[i])!=null){                int count = hashMap_s.get(char_s[i]);                hashMap_s.put(char_s[i], ++count);            }else{                hashMap_s.put(char_s[i], 1);            }        }        for(int i=0;i<char_t.length;i++){            if(hashMap_s.get(char_t[i])!=null){                int count = hashMap_s.get(char_t[i]);                if(count>1){                    hashMap_s.put(char_t[i], --count);                }else{                    hashMap_s.remove(char_t[i]);                }            }else{                isAnagram = false;                break;            }        }        return isAnagram;    }}

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