有人可以向我解释下面的标记线是如何工作的吗？他们到底做了什么？
public class GrammerUtils {
public static void main(String args[]) {
System.out.print(isAnagram("teacher", "cheater"));
}
public static boolean isAnagram(String word, String anagram) {
if (word.length() != anagram.length()) {
return false;
}
char[] chars = word.toCharArray(); // marked
for (char c: chars) { // marked
int index = anagram.indexOf(c);// marked
if (index != -1) { // marked
anagram = anagram.substring(0, index) + anagram.substring(index + 1, anagram.length());
} else {
return false;
}
}
return anagram.isEmpty();
}
}
答案 0 :(得分：3)
代码在单词之后接受一个字符
char[] chars = word.toCharArray();
for (char c: chars) {
并检查它是否在潜在的字谜中。
int index = anagram.indexOf(c);
如果是的话
if (index != -1) {
然后删除它，以便在重复时不再检查它（你在字符之前的子字符串和字符后面的子字符串，请记住子字符串在第二个参数中是独占的）：
anagram = anagram.substring(0, index) + anagram.substring(index + 1, anagram.length());
并转到下一个字符（当您完成检查单词中的所有字符时，检查字谜中的所有字符是否匹配，即anagram现在是否为空）。
如果角色不在字谜中，则表示它不是字谜，并且您返回false。
return false;
这是相当低效的（O（n ^ 2））。最好对两个字符串进行排序并比较结果（O（n * log（n））
答案 1 :(得分：1)
我在不使用索引的情况下尝试了以下方式
package com.learn.java.fileop;
public class Anagram {
public boolean checkAnagram(String basestr , String anastr){
char[] ch = basestr.toCharArray();
char[] ch1 = anastr.toCharArray();
int asciivalue=0;
int sum_src=0;
for(int i =0 ; i< ch.length;i++){
asciivalue = ch[i];
sum_src = sum_src+asciivalue;
}
System.out.println(""+sum_src);
int sum_dest=0;
for(int i = 0 ; i< ch1.length;i++){
asciivalue = ch1[i];
sum_dest = sum_dest+asciivalue;
}
System.out.println(""+sum_dest);
if (sum_src == sum_dest){
return true;
}else{
return false;
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Anagram an = new Anagram();
System.out.println(an.checkAnagram("teacher", "cheater"));
}
}
如果这看起来不错或有任何其他想法，请告诉我。
答案 2 :(得分：1)
这不能回答你的问题，但渐渐地是一种更有效的算法。
public static void isAnagram(String s1, String s2){
char[] c1 = s1.toLowerCase().toCharArray();
char[] c2 = s2.toLowerCase().toCharArray();
Arrays.sort(c1);
Arrays.sort(c2);
if(Arrays.equals(c1, c2))
System.out.println("s1 is anagram of s2");
else
System.out.println("Strings are not anagram");
}
答案 3 :(得分：0)
代码循环遍历word的每个字符。对于每个字符，它会检查它是否在anagram中。如果不是，则返回false。否则，它会从anagram中删除该字符，然后转到下一个字符。
public static boolean isAnagram(String word, String anagram) {
if (word.length() != anagram.length()) {
return false;
}
char[] chars = word.toCharArray();
//loop through each character in `word`
for (char c: chars) {
int index = anagram.indexOf(c);
//if it exists in `anagram`, remove it using a combination of `substring` calls, else return false
if (index != -1) {
anagram = anagram.substring(0, index) + anagram.substring(index + 1, anagram.length());
} else {
return false;
}
}
return anagram.isEmpty();
}
答案 4 :(得分：0)
代码的效率相当低，因为它在for循环的每一步都创建了一个新的String，并将其重新分配给 anagram 。下面是一种使用java.util.Arrays的方法，它包含用于操作数组的各种实用程序方法。最后，isAnagarm（）可能应该是一个静态的util方法。
public static boolean isAnagram(String s, String t) {
if (s == null || t == null) {
return false;
}
if(s.isEmpty() && t.isEmpty()) {
return true;
}
char[] sArray = s.toCharArray();
char[] tArray = t.toCharArray();
Arrays.sort(sArray);
Arrays.sort(tArray);
return Arrays.equals(sArray, tArray);
}
答案 5 :(得分：0)
Program to check the entered strings are Anagram:
public class AnagramWordTest {
public static void main(String[] args) {
String str1 = "cat";
String str2 = "tac";
int not_found=0;
if(str1.length() == str2.length()) {
Boolean isDuplicate =
testDuplicatesinEachWord(str1.toLowerCase().trim(),
str2.toLowerCase().trim());
if(!isDuplicate) {
int found=0;
for (int i = 0; i < str1.length(); i++) {
for (int j = 0; j < str2.length(); j++) {
if(str1.charAt(i) == str2.charAt(j)) {
found=1;
break;
}
}
if(found == 0) {
not_found=1;
break;
}
}
if(not_found==1) {
System.out.println("The two strings are not Anagrams");
}else {
System.out.println("The two strings are Anagrams");
}
} else {
System.out.println("Entered strings has duplicates to check
Anagrams in either");
}
}else {
System.out.println("String lengths are different and are not
Anagram");
}
}
private static boolean testDuplicatesinEachWord(String str1, String str2)
{
Boolean isStr1Duplicate = false;
Boolean isStr2Duplicate = false;
Boolean isDuplicate = false;
for(int k=0;k<str1.length();k++) {
for(int m=k+1;m<str1.length()-1;m++) {
if(str1.charAt(k) == str1.charAt(m)) {
isStr1Duplicate = true;
break;
}
}
}
for(int l=0;l<str2.length()-1;l++) {
for(int n=l+1;n<str1.length()-1;n++) {
if(str1.charAt(l) == str2.charAt(n)) {
isStr2Duplicate = true;
break;
}}
}
if(isStr1Duplicate.equals(Boolean.TRUE) ||
isStr2Duplicate.equals(Boolean.TRUE)) {
isDuplicate= true;
}
return isDuplicate;
}
}
答案 6 :(得分：-1)
//导入将帮助我们操纵数组的Arrays类。 导入java.util.Arrays;
公共类AnagramTest {
private static boolean isAnagram(String str1 , String str2){
// Converting both strings to char arrays as strings do not have direct
// sorting method in java.
char [] leftArray = ( str1.trim().toLowerCase()).toCharArray();
char [] rightArray = ( str2.trim().toLowerCase()).toCharArray();
Arrays.parallelSort(leftArray); // Sorting the array using the java 8
Arrays.parallelSort(rightArray);// parallelSort method
return Arrays.equals(leftArray, rightArray);
}
public static void main(String [] args){
//Test cases;
String a = "integral"; // initializing first string
String b = "Triangle"; // initializing second string
System.out.println("The statement "+ a + " is anagram of "+b+" is "+isAnagram(a,b));// Print true
String c = "silent"; // initializing first string
String d = "listen"; // initializing second string
System.out.println("The statement "+ c + " is anagram of "+d+" is "+isAnagram(c,d));// Print true
String e = "fried"; // initializing first string
String f = "fired"; // initializing second string
System.out.println("The statement "+ e + " is anagram of "+f+" is "+isAnagram(e,f));// true
String g = "ball"; // initializing first string
String h = "call"; // initializing second string
System.out.println("The statement "+ g + " is anagram of "+h+"is "+isAnagram(g,h));// Print false
}
}