当我以这种方式定义我的数组字符串时:
String[] X = {"X","M","J","Y","A","U","Z"};
String[] Y = {"M","Z","J","A","W","X","U"};
我的代码有效,它打印
[M, J, A, U]
这是X
的最长公共(public)子序列和 Y
但是当我为具有相同输入的字符串数组定义文本文件时,我的代码会打印一个空数组 []
.我该如何解决这个问题? 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;
}
}
最佳答案
您应该使用 Object#equals(Object anotherObject)
方法来比较字符串,或者一般来说,每个对象。
在您的代码中,您使用的是 ==
运算符将比较字符串引用(如果它们是相同的 String
实例)而不是它们的值。
您的代码有效(使用 Array 初始化程序时),因为您初始化了 String
具有文字和两个相同文字的数组将是同一个实例。
当你用 readLine()
读取文件中的一行时, 它创建一个新的 String
所以具有相同内容的两行将导致两个具有相同值但不同实例的字符串。
因此,在比较字符串时,请使用 equals
并且您的代码将起作用。
另见:What is the difference between == and equals() in Java?
关于java - 为什么我的代码在使用文本文件时打印空数组([])?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59345581/