我有一个字符串,名为"abab“,我把它分成两对(即ab,ab)。如果pair已经存在,那么我不希望它是generated.How。我这样做了吗?
String r="abab";
String pair[] = new String[r.length()/2];
for( int i = 0; i <pair.length; i++ )
{
pair[i] = r.substring(i*2,(i*2)+2);
}
发布于 2016-04-09 00:50:02
在将它添加到对数组之前,可以使用Arrays函数.contains查看它是否已经存在。如果配对已经存在,则不要将其添加到配对列表中。例如,这里不会添加ab和fe对:
String r="ababtefedefe";
String pair[] = new String[r.length()/2];
String currentPair = "";
for( int i = 0; i <pair.length; i++ )
{
currentPair = r.substring(i*2,(i*2)+2);
if(!java.util.Arrays.asList(pair).contains(currentPair))
pair[i] = currentPair;
System.out.println(pair[i]);
}
发布于 2016-04-09 01:01:38
我会用一套来帮我脱身。
private String[] retrieveUniquePair(String input) {
int dim = input.length() / 2;
Set<String> pairs = new LinkedHashSet<>(dim);
for (int i = 0; i <= dim; i += 2) {
String currentPair = input.substring(i, i + 2);
pairs.add(currentPair);
}
return pairs.toArray(new String[] {});
}
编辑:我发布了我提出的解决方案和测试
public class PairTest {
@DataProvider(name = "input")
public static Object[][] input() {
return new Object[][] {
{"abcd", Arrays.asList("ab", "cd")},
{"abcde", Arrays.asList("ab", "cd")},
{"abcdab", Arrays.asList("ab", "cd")},
{"ababcdcd", Arrays.asList("ab", "cd")},
{"ababtefedefe", Arrays.asList("ab", "te", "fe", "de")},
};
}
@Test(dataProvider = "input")
public void test(String input, List<String> expectedOutput) {
String[] output = retrieveUniquePair(input);
Assert.assertNotNull(output);
Assert.assertEquals(output.length, expectedOutput.size());
for (String pair : output) {
Assert.assertTrue(expectedOutput.contains(pair));
}
}
private String[] retrieveUniquePair(String input) {
int pairNumber = input.length() / 2;
Set<String> pairs = new LinkedHashSet<>(pairNumber);
int endIteration = input.length();
if (input.length() % 2 != 0) { // odd number
endIteration--; // ignore last character
}
for (int i = 0; i < endIteration; i += 2) {
String currentPair = input.substring(i, i + 2);
pairs.add(currentPair);
}
return pairs.toArray(new String[pairs.size() - 1]);
}
}
https://stackoverflow.com/questions/36504678
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