针对string读取错误的方法（+anagram题解）_string识别错误

Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are anagram of each other or not. An anagram of a string is another string that contains same characters, only the order of characters can be different. For example, “act” and “tac” are anagram of each other.

Input:
The first line of input contains an integer T denoting the number of test cases. Each test case consist of two strings in 'lowercase' only, in a single line.

Output:
Print "YES" without quotes if the two strings are anagram else print "NO".

Constraints:
1 ≤ T ≤ 300
1 ≤ |s| ≤ 106

Example:
Input:
2
geeksforgeeks forgeeksgeeks
allergy allergic

Output:
YES
NO

Explanation:
Testcase 1: Both the string have same characters with same frequency. So, both are anagrams.
Testcase 2: Characters in both the strings are not same, so they are not anagrams.
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/*package whatever //do not write package name here */

import java.util.*;
import java.lang.*;
import java.io.*;

class GFG {
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine()); //2
		while(t-->0) {
			String str = br.readLine();  //读取一行字符串 geeksforgeeks forgeeksgeeks
			String s1 = str.split(" ")[0];  //返回空格分隔的1th string：geeksforgeeks
			String s2 = str.split(" ")[1];  //返回空格分隔的2nd string：forgeeksgeeks
			char[] str1 = s1.toCharArray(); //转换为char数组
			char[] str2 = s2.toCharArray();  //转换为char数组
			
			if(anagram(str1,str2)) {
				System.out.println("YES");
			} else {
				System.out.println("NO");
			}
		}

	}
		private static boolean anagram(char[] str1, char[] str2) {
		// TODO Auto-generated method stub
		int n1 = str1.length;
		int n2 = str2.length;
		
		if(n1 != n2) return false;
		
		Arrays.sort(str1);
		Arrays.sort(str2);
		
		for(int i = 0; i < n1; i++) {
			if(str1[i] != str2[i]) return false;
		}
		return true;
	}
}
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