WordPattern

Given a pattern and a string str, find if str follows the same pattern.

Examples:

1. pattern = "abba", str = "dog cat cat dog" should return true.
2. pattern = "abba", str = "dog cat cat fish" should return false.
3. pattern = "aaaa", str = "dog cat cat dog" should return false.
4. pattern = "abba", str = "dog dog dog dog" should return false.

Notes:

1. patterncontains only lowercase alphabetical letters, and str contains words separated by a single space. Each word in str contains only lowercase alphabetical letters.
2. Both pattern and str do not have leading or trailing spaces.
3. Each letter in pattern must map to a word with length that is at least 1.

在LeeCode上有更简便的方法，使用了Map我对map还不够熟悉，所以在解决这个问题的时候，没有想起来用它

代码：

import java.util.Arrays;

public class WordPattern {

public static void main(String[] args) {
String pattern = "abab";
String str = "abc qwe abc qwe";
boolean compare = wordPattern(pattern, str);
if (compare) {
System.out.println("true");
} else {
System.out.println("false");
}
}

// 模式匹配
public static boolean wordPattern(String pattern, String str) {
// 转换为数组
String[] strings = str.split(" ");
char[] strings2 = pattern.toCharArray();

// 判断长度
if (strings2.length != strings.length) {
return false;
}

// 数组记录替换
int[] result1 = replaceString(strings);
int[] result2 = replaceChar(strings2);
Arrays.sort(result1);
Arrays.sort(result2);
if (!Arrays.equals(result1, result2)) {
return false;
}
return true;
}

// 数组替换
private static int[] replaceChar(char[] strings2) {
int[] array = new int[strings2.length];
for (int i = 0; i < array.length; i++) {
array[i] = 0;
}
for (int i = 0; i < strings2.length; i++) {
for (int j = 0; j < strings2.length; j++) {
char temp = strings2[i];
if (strings2[j] == temp) {
if (array[j] == 0 && temp == strings2[j]) {
array[j] = i + 1;
}
}
}
}
return array;
}

// 数组替换
private static int[] replaceString(String[] strings) {
int[] array = new int[strings.length];
for (int i = 0; i < array.length; i++) {
array[i] = 0;
}
for (int i = 0; i < strings.length; i++) {
String temp = strings[i];
for (int j = 0; j < strings.length; j++) {
if (temp.equals(strings[j]) && array[j] == 0) { // 此处注意 == 和
// equals的区别
array[j] = i + 1;
}
}
}
return array;
}

}