[英]How to find longest common suffix from a list of strings and return the resulting suffix length in java?
请在ArrayList或LinkedList中考虑以下内容:[Gloucestershire,Hampshire,Yorkshire,Lancashire] shire是长度为5的最长的通用后缀。输出应为5我如何编写一种方法来实现上述和返回长度
Try this. 尝试这个。
Explanation is in the comments 注释中有解释
package javaapplication1;
public class SegmentTree {
public static void main(String[] args) {
String[] array = {"Gloucestershire", "Hampshire", "Yorkshire", "Lancashire"};
int i,j;
int min=10000000;
//reversing the strings and finding the length of smallest string
for(i=0;i<(array.length);i++)
{
if(array[i].length()<min)
min=array[i].length();
StringBuilder input1 = new StringBuilder();
input1.append(array[i]);
array[i] = input1.reverse().toString();
}
//finding the length of longest suffix
for(i=0;i<min;i++)
{
for(j=1;j<(array.length);j++)
{
if(array[j].charAt(i)!=array[j-1].charAt(i))
{
break;
}
}
if(j!=array.length)
break;
}
System.out.println(i);
}
}
What I am doing here is,first checking the last elem of all strings, then 2nd last and so on. 我在这里所做的是,首先检查所有字符串的最后一个元素,然后检查第二个等等。
Time Complexity: O(n*m) , 时间复杂度:O(n * m) ,
n=number of strings, m=length of smallest string n =字符串数,m =最小字符串的长度
Guava has a helper function called Strings.commonSuffix(CharSequence a, CharSequence b)
that could be used in your algorithm. 番石榴有一个名为
Strings.commonSuffix(CharSequence a, CharSequence b)
的辅助函数,可以在您的算法中使用。 I know adding dependency like Guava only to have this function can be overkill - in this case you can take a look at the source code to see how this function is implemented and you can move this implementation to your program. 我知道添加像Guava这样的依赖项仅具有该功能可能会过大-在这种情况下,您可以查看源代码以了解如何实现此功能,并将此实现移至程序中。 Then your program could look like this:
然后您的程序可能如下所示:
import com.google.common.base.Strings;
import java.io.IOException;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class CommonSuffixLengthMain {
public static void main(String[] args) throws IOException {
assert 5 == commonSuffixLength(Arrays.asList("Gloucestershire", "Hampshire", "Yorkshire", "Lancashire"));
assert 2 == commonSuffixLength(Arrays.asList("abc", "dbc", "qebc", "webc"));
assert 0 == commonSuffixLength(Collections.emptyList());
assert 0 == commonSuffixLength(Collections.singletonList("abc"));
assert 0 == commonSuffixLength(Arrays.asList("abc", "def", "zxc"));
}
private static int commonSuffixLength(final List<String> strings) {
int result = 0;
if (strings == null || strings.size() < 2) {
return result;
}
for (int i = 0; i < strings.size() - 1; i++) {
String prefix = Strings.commonSuffix(strings.get(i), strings.get(i + 1));
result = result == 0 ?
prefix.length() :
Math.min(prefix.length(), result);
}
return result;
}
}
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