Given a pattern and a string str, find if str follows the same pattern.
Examples:
- pattern =
"abba", str ="dog cat cat dog"should return true. - pattern =
"abba", str ="dog cat cat fish"should return false. - pattern =
"aaaa", str ="dog cat cat dog"should return false. - pattern =
"abba", str ="dog dog dog dog"should return false.
Notes:
-
patterncontains only lowercase alphabetical letters, andstrcontains words separated by a single space. Each word instrcontains only lowercase alphabetical letters. - Both
patternandstrdo not have leading or trailing spaces. - Each letter in
patternmust 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;
}
}