java 检查一个字符串是否包含另一个字符串的所有字符
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Check if a string has all the characters of another string
提问by Toby Mellor
String one = "This is a test";
String two = "This is a simple test";
I want to check if twocontains all the characters that are in one, and ignore the fact it has extra characters.
我想检查是否two包含 中的所有字符one，并忽略它有多余字符的事实。
回答by Mureinik
The fastest thing would probably be to break them up to HashSets and then apply containsAll
最快的方法可能是将它们分解为HashSets 然后应用containsAll
public static Set<Character> stringToCharacterSet(String s) {
Set<Character> set = new HashSet<>();
for (char c : s.toCharArray()) {
set.add(c);
}
return set;
}
public static boolean containsAllChars
(String container, String containee) {
return stringToCharacterSet(container).containsAll
(stringToCharacterSet(containee));
}
public static void main(String[] args) {
String one = "This is a test";
String two = "This is a simple test";
System.out.println (containsAllChars(one, two));
}
回答by Paul Boddington
static boolean stringContains(String longer, String shorter) {
int i = 0;
for (char c : shorter.toCharArray()) {
i = longer.indexOf(c, i) + 1;
if (i <= 0) { return false; }
}
return true;
}
回答by M Anouti
Using a simple loop over the set of the characters in the first string:
在第一个字符串中的字符集上使用一个简单的循环：
String s1 = "This is a test";
String s2 = "This is a simple test";
Set<Character> chars = new HashSet<Character>();
for(int i = 0; i < s1.length(); i++) {
chars.add(s1.charAt(i));
}
for (Iterator<Character> iterator = chars.iterator(); iterator.hasNext();) {
Character character = iterator.next();
if(!s2.contains(character.toString())) {
// break and mark as not contained
break;
}
}
If words are meant to be checked, then you could split the string around whitespaces into a list of words:
如果要检查单词，则可以将空格周围的字符串拆分为单词列表：
String[] words1 = s1.split("\s");
String[] words2 = s2.split("\s");
List<String> wordList1 = Arrays.asList(words1);
List<String> wordList2 = Arrays.asList(words2);
System.out.println(wordList2.containsAll(wordList1));
回答by harold_finch
Try this. I know it's long but it works
试试这个。我知道它很长但它有效
public static void main(String[] args)
{
String String1,String2;
int i, j, count1, count2;
count1 = 0;
count2 = 0;
String1 = "This is a test";
String2 = "This is a simple test";
char[] list1 = new char[String2.length()];
char[] list2 = new char[String2.length()];
char[] list3 = new char[String2.length()];
for (i = 0; i <= String1.length() - 1; i++)
{
list1[i] = String1.charAt(i);
for (j = 0; j <= String2.length() - 1; j++)
{
list2[j] = String2.charAt(j);
if (list1[i] == list2[j])
{
i++;
count1++;
}
}
}
for (i = 0; i <= String1.length() - 1; i++)
{
list1[i] = String1.charAt(i);
for (j = 0; j <= String1.length() - 1; j++)
{
list3[j] = String1.charAt(j);
if (list1[i] == list3[j])
{
i++;
count2++;
}
}
}
if (count1 >= count2)
System.out.println(true);
else
System.out.println(false);
}
回答by justanother
two.startsWith(one)
If you aren't sure about the start position (the above assumes 0), try using the following API
如果您不确定起始位置（以上假设为 0），请尝试使用以下 API
startsWith(String, offset)
