java需要一个递归算法来确定一个字符串是否是其他两个字符串的混洗
我试图创建一个递归算法来确定字符串c是否是字符串a和b的有序无序排列。有序无序排列意味着字符串c是通过以一种仍然保持两个字符串从左到右顺序的方式穿插字符串a和b的字符组成的
我已经尝试了这个算法的代码,它在某种程度上是有效的,但是当我尝试测试某个混乱的单词时遇到了一个问题,正如我下面的代码所示。我相信这与我的if语句中的索引错误有关,但是无法找到修复此语句或绕过此语句的方法,任何反馈或指导都将非常感谢
public class StringShuffleTest {
public static boolean isOrderedShuffle(String a, String b, String c){
//boolean to determine if String c is an ordered shuffle.
boolean isShuffled = false;
//variables for the size of Strings a, b and c.
int n = a.length();
int m = b.length();
int len = c.length();
//if the length of c is not the length of a + b return false.
if (len != (n + m)){
return isShuffled;
}
//if the length of a or b is 0, and c equals a or b, return true, otherwise,
//return false.
if (n == 0 || m == 0){
if (c.equals(a) || c.equals(b)){
return true;
}
else
return isShuffled;
}
//if String a has length 1, remove String a from String c and make String a empty.
if (n == 1){
c = c.substring(0, c.indexOf(a.charAt(0))) + c.substring(c.indexOf(a.charAt(0)) +1);
a = "";
return isOrderedShuffle(a, b, c);
}
else
//Recursive algorithm to determine if String c is an ordered shuffle of a and b.
if (c.indexOf(a.charAt(0)) >= 0){
int indexOfFirst = c.indexOf(a.charAt(0));
int indexOfSecond = c.indexOf(a.charAt(1));
if (indexOfFirst <= indexOfSecond){
c = c.substring(0, indexOfFirst) + c.substring(indexOfFirst +1);
a = a.substring(1, n);
System.out.println(a);
System.out.println(c);
return isOrderedShuffle(a, b, c);
}
else
return isShuffled;
}
return isShuffled;
}
public static void main(String[] args) {
System.out.println(StringShuffleTest.isOrderedShuffle("castle", "cat", "catcastle"));
}
}
# 1 楼答案
您可以通过创建一个调用递归方法的方法来简化此过程,例如: