我是计算算法的运行时复杂度的新手，并且通常在算法中有for循环时用于计算复杂性，并且不确定如何正确计算算法的复杂度，如它主要只是许多if语句的列表。
代码可以在下面看到，算法测试以查看String c是否是字符串a和b的有序shuffle。
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 String c contains String b as a substring, then remove String b from c and make m = 0.
if (c.contains(b)){
c = c.replace(b, "");
m = 0;
}
//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 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);
}
//An ordered shuffle of two given strings, a and b, is a string that can be formed by interspersing
//the characters of a and b in a way that maintains the left-to-right order of the characters from each
//string.
//Recursive algorithm to determine if String c is an ordered shuffle of a and b.
else
if (c.indexOf(a.charAt(0)) >= 0){
int indexOfFirsta = c.indexOf(a.charAt(0));
int indexOfSeconda = c.indexOf(a.charAt(1));
if (indexOfFirsta <= indexOfSeconda){
c = c.substring(0, indexOfFirsta) + c.substring(indexOfFirsta +1);
a = a.substring(1, n);
System.out.println(a);
System.out.println(c);
return isOrderedShuffle(a, b, c);
}
else
if (c.indexOf(b.charAt(0)) >= 0){
int indexOfFirstb = c.indexOf(b.charAt(0));
int indexOfSecondb = c.indexOf(b.charAt(1));
if (indexOfFirstb <= indexOfSecondb){
c = c.substring(0, indexOfFirstb) + c.substring(indexOfFirstb +1);
b = b.substring(1, m);
System.out.println(b);
System.out.println(c);
return isOrderedShuffle(a, b, c);
}
}
}
return isShuffled;
}
public static void main(String[] args) {
System.out.println(StringShuffleTest.isOrderedShuffle("cat", "castle", "castlecat"));
}
}
如果c的第一个字母不等于a或b的第一个字母，最初的情况应该是t（n），我认为最好的运行时间复杂度应该是t（n）= 1 = n + m，如果必须比较String c中每个字母的字符。任何帮助或建议将不胜感激。