当我以此方式定义数组字符串时,“ String [] X = {” X“,” M“,” J“,” Y“,” A“,” U“,” Z“}},Y = { “ M”,“ Z”,“ J”,“ A”,“ W”,“ X”,“ U”}“我的代码有效,并且它打印最长的” [M,J,A,U]“ X和Y的常见子序列,但是当我为具有相同输入的字符串数组定义文本文件时,我的代码将打印一个空数组“ []”。我该如何解决?
public class LCS {
// Function to find LCS of String X[0..m-1] and Y[0..n-1]
public static String A(String[] x, String[] y, int m, int n, int[][] T)
{
// return empty string if we have reached the end of
// either sequence
if (m == 0 || n == 0) {
return new String();
}
// if last character of X and Y matches
if (x[m - 1] == y[n - 1])
{
// append current character (X[m-1] or Y[n-1]) to LCS of
// substring X[0..m-2] and Y[0..n-2]
return A(x, y, m - 1, n - 1, T) + x[m - 1];
}
// else when the last character of X and Y are different
// if top cell of current cell has more value than the left
// cell, then drop current character of String X and find LCS
// of substring X[0..m-2], Y[0..n-1]
if (T[m - 1][n] > T[m][n - 1]) {
return A(x, y, m - 1, n, T);
}
else {
// if left cell of current cell has more value than the top
// cell, then drop current character of String Y and find LCS
// of substring X[0..m-1], Y[0..n-2]
return A(x, y, m, n - 1, T);
}
}
// Function to fill lookup table by finding the length of LCS
// of substring X[0..m-1] and Y[0..n-1]
public static void LCSLength(String[] x, String[] y, int m, int n, int[][] T)
{
// fill the lookup table in bottom-up manner
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
// if current character of X and Y matches
if (x[i - 1] == y[j - 1]) {
T[i][j] = T[i - 1][j - 1] + 1;
}
// else if current character of X and Y don't match
else {
T[i][j] = Integer.max(T[i - 1][j], T[i][j - 1]);
}
}
}
}
// main function
public static void main(String[] args) throws IOException
{
String[] X = read("C:\\Users\\fener\\Desktop\\producerconsumer\\Yeni Metin Belgesi.txt");
String[] Y = read("C:\\Users\\fener\\Desktop\\producerconsumer\\Yeni Metin Belgesi (2).txt");
//String[] X = {"X","M","J","Y","A","U","Z"}, Y = {"M","Z","J","A","W","X","U"};
int m = X.length, n = Y.length;
// T[i][j] stores the length of LCS of substring
// X[0..i-1], Y[0..j-1]
int[][] T = new int[m + 1][n + 1];
// fill lookup table
LCSLength(X, Y, m, n, T);
String[] arr = A(X, Y, m, n, T).split("");
// find longest common sequence
System.out.print(Arrays.toString(arr));
System.exit(0);
}
private static String[] read(String location) throws IOException {
BufferedReader reader1 = new BufferedReader(new FileReader(location));
String line;
ArrayList<String> lines = new ArrayList<String>();
while ((line = reader1.readLine()) != null) {
lines.add(line);
}
reader1.close();
String[] result = new String[lines.size()];
for(int i=0; i<lines.size(); i++) {
result[i] = lines.get(i);
}
return result;
}
}
""
和使用new String()
构造函数实例化字符串是有区别的。例如:
// Example 1
String a = "Y";
String b = "Y";
boolean result1 = a == b; // true
// Example 2
String c = new String("Y");
String d = new String("Y");
boolean result2 = c == d; // false
这是因为当您使用"Y"
创建字符串时,实际对象分配在堆中称为字符串常量池的单独位置。 "Y"
的任何后续分配都将返回对字符串常量池中相同对象的引用。
[当您使用new String("Y")
时,您要在公共堆中分配一个全新的String对象实例。
==
运算符比较2个对象以确定它们是否引用相同的对象实例,在这种情况下,该实例将与示例2所示的实例不同。因此,对于此代码,我们只需要更改:
// return empty string if we have reached the end of // either sequence if (m == 0 || n == 0) { return ""; } // if last character of X and Y matches if (Objects.equals(x[m - 1], y[n - 1])) { ... // if current character of X and Y matches if (Objects.equals(x[i - 1], y[j - 1])) { ...
这里java.util.Objects.equals
安全地与==
比较,然后与equals()
比较。
通过应用这些更改,结果是:
[M, J, A, U]
最后,可以使用read
API简化java.nio
方法:
private static String[] read(String folder, String filename) throws IOException {
Path path = Paths.get(folder, filename);
List<String> lines = Files.readAllLines(path);
return lines.toArray(new String[0]);
}