我想不出比暴力解决方案更好的了:尝试单词的每种可能的分割(这本质上就是Jan所描述的)。
运行时间复杂度为O(n^(2m))
,其中m
是模式的长度,n
是字符串的长度。
以下是代码示例(我让我的代码返回实际映射而不仅仅是0或1。将代码修改为返回0或1很容易):
import java.util.Arrays;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class StringBijection {
public static void main(String[] args) {
String chars = "abaac";
String string = "johnjohnnyjohnjohncodes";
List<String> stringBijection = getStringBijection(chars, string);
System.out.println(Arrays.toString(stringBijection.toArray()));
}
public static List<String> getStringBijection(String chars, String string) {
if (chars == null || string == null) {
return null;
}
Map<Character, String> bijection = new HashMap<Character, String>();
Deque<String> assignments = new ArrayDeque<String>();
List<String> results = new ArrayList<String>();
boolean hasBijection = getStringBijection(chars, string, 0, 0, bijection, assignments);
if (!hasBijection) {
return null;
}
for (String result : assignments) {
results.add(result);
}
return results;
}
private static boolean getStringBijection(String chars, String string, int charIndex, int stringIndex, Map<Character, String> bijection, Deque<String> assignments) {
int charsLen = chars.length();
int stringLen = string.length();
if (charIndex == charsLen && stringIndex == stringLen) {
return true;
} else if (charIndex == charsLen || stringIndex == stringLen) {
return false;
}
char currentChar = chars.charAt(charIndex);
List<String> possibleWords = new ArrayList<String>();
boolean charAlreadyAssigned = bijection.containsKey(currentChar);
if (charAlreadyAssigned) {
String word = bijection.get(currentChar);
possibleWords.add(word);
} else {
StringBuilder word = new StringBuilder();
for (int i = stringIndex; i < stringLen; ++i) {
word.append(string.charAt(i));
possibleWords.add(word.toString());
}
}
for (String word : possibleWords) {
int wordLen = word.length();
int endIndex = stringIndex + wordLen;
if (endIndex <= stringLen && string.substring(stringIndex, endIndex).equals(word)) {
if (!charAlreadyAssigned) {
bijection.put(currentChar, word);
}
assignments.addLast(word);
boolean done = getStringBijection(chars, string, charIndex + 1, stringIndex + wordLen, bijection, assignments);
if (done) {
return true;
}
assignments.removeLast();
if (!charAlreadyAssigned) {
bijection.remove(currentChar);
}
}
}
return false;
}
}