Java中的str.contains
我有两张名单Java中的str.contains,java,string,string-matching,contain,Java,String,String Matching,Contain,我有两张名单 allWords [book, pen, pencil] subsetString [book pen , book pencil , pen pencil ] 我希望我的输出为 book pen => pencil book pencil => pen pen pencil => book 对于子字符串中的每个元素,我将使用allwords进行检查。 一旦未找到匹配项,则allwords中的字符串将添加到输出的RHS中 但我的问题是 现在
allWords [book, pen, pencil]
subsetString [book pen , book pencil , pen pencil ]
我希望我的输出为
book pen => pencil
book pencil => pen
pen pencil => book
对于子字符串中的每个元素,我将使用allwords进行检查。
一旦未找到匹配项,则allwords中的字符串将添加到输出的RHS中
但我的问题是
现在我只得到2个输出,而不是3个
allWords [book, pen, pencil]
subsetString [book pen , book pencil , pen pencil ]
pen pencil => book
book pen => pencil
原因是当进入book pencil
时,它会被allWords[book,pen,pencil]
一旦涉及到pen包含书本铅笔
,它就会感到满意(因为铅笔
包含子字符串笔
)
代码
public void getStrongRules2(ArrayList<String> allWords,String delim) {
ArrayList<ArrayList<String>> subsets = BTSum(allWords);
ArrayList<String> subsetString = new ArrayList<String>();
for (int j = 0; j < subsets.size(); j++) {
String substring = "";
for (int k = 0; k < subsets.get(j).size(); k++) {
substring += subsets.get(j).get(k) + " ";
}
subsetString.add(substring);
}
System.out.println("allWords "+allWords);
System.out.println("subsetString "+subsetString);
for(String a : allWords){
for (int j = 0; j < subsetString.size(); j++) {
if (!(subsetString.get(j).contains(a))) {
System.out.println(subsetString.get(j)+" => "+a);
}
}
}
}
public static ArrayList<ArrayList<String>> BTSum(ArrayList<String> numbers) {
int n = numbers.size();
ArrayList<ArrayList<String>> powerSet = new ArrayList<ArrayList<String>>();
for (long i = 0; i < (1 << n); i++) {
ArrayList<String> element = new ArrayList<String>();
for (int j = 0; j < n; j++)
if ((i >> j) % 2 == 1) {
element.add(numbers.get(j));
}
if (element.size() > 1 && element.size() < n) {
powerSet.add(element);
}
}
return powerSet;
}
}
public void getStrongRules2(ArrayList allWords,String delim){
ArrayList子集=BTSum(所有字);
ArrayList subsetString=新的ArrayList();
对于(int j=0;j”+a);
}
}
}
}
公共静态ArrayList BTSum(ArrayList编号){
int n=numbers.size();
ArrayList powerSet=新的ArrayList();
对于(长i=0;i<(1>j)%2==1){
元素.add(number.get(j));
}
if(element.size()>1&&element.size()
但这不应该发生在我身上
如何纠正这一点
请建议只需按长度desc对子字符串进行排序,然后查找它们。 使用自定义压缩机,如:
public class MyComparator implements java.util.Comparator<String> {
private int referenceLength;
public MyComparator(String reference) {
super();
this.referenceLength = reference.length();
}
public int compare(String s1, String s2) {
int dist1 = Math.abs(s1.length() - referenceLength);
int dist2 = Math.abs(s2.length() - referenceLength);
return dist1 - dist2;
}
}
这是因为(subsetString.get(j).contains(a))->“pen pencil.contains('pen')是真的。因此,它不会进入if循环内部打印结果。问题是
书本铅笔'pen'cil包含pen
,因此使用contains()
无法获得正确的结果
所以你可以试试下面的方法
ArrayList<String> allWords = new ArrayList<String>() {{
add("book");
add("pen");
add("pencil");
}};
ArrayList<String> subsetString = new ArrayList<String>() {{
add("book pen");
add("book pencil");
add("pen pencil");
}};
for (String i : subsetString) {
List<String> stringList = Arrays.asList(i.split(" "));
for (String j : allWords) {
if(!stringList.contains(j)) {
System.out.println(i + " => " + j);
}
}
}
ArrayList allWords=new ArrayList(){{
添加(“账簿”);
添加(“笔”);
加上(“铅笔”);
}};
ArrayList subsetString=新的ArrayList(){{
添加(“书笔”);
添加(“书本铅笔”);
添加(“钢笔”);
}};
用于(字符串i:子字符串){
List stringList=Arrays.asList(i.split(“”);
for(字符串j:allWords){
如果(!stringList.contains(j)){
系统输出println(i+“=>”+j);
}
}
}
您不需要所有这些代码。只需使用以下工具即可完成所有重物搬运:
List<String> allWords = Arrays.asList("book", "pen", "pencil");
List<List<String>> subsetString = Arrays.asList(
Arrays.asList("book", "pen"),
Arrays.asList("book", "pencil"),
Arrays.asList("pen", "pencil"));
for (List<String> subset : subsetString) {
ArrayList<String> strings = new ArrayList<String>(allWords);
strings.removeAll(subset);
System.out.println(subset + " => " + strings);
}
您可以这样做:
List<String> allWords = new ArrayList<String>();
allWords.add("book");
allWords.add("pen");
allWords.add("pencil");
List<String> subsetStrring = new ArrayList<String>();
subsetStrring.add("book pen");
subsetStrring.add("book pencil");
subsetStrring.add("pen pencil");
for (String string : subsetStrring) {
List<String> subsetStr = Arrays.asList(string.split(" "));
for (String word : allWords) {
if(!subsetStr.contains(word)) {
System.out.println(string + " => " + word);
break;
}
}
}
List allWords=new ArrayList();
所有词语。添加(“书籍”);
所有文字。添加(“笔”);
添加(“铅笔”);
List subsettrring=new ArrayList();
子条款添加(“书笔”);
子条款添加(“书笔”);
子条款添加(“钢笔”);
用于(字符串:子字符串){
List subsettr=Arrays.asList(string.split(“”));
for(字符串字:allWords){
如果(!subsettr.包含(word)){
System.out.println(字符串+“=>”+字);
打破
}
}
}
考虑
List<String> allWords = new ArrayList<>();
allWords.add("Book"); allWords.add("Pen"); allWords.add("Pencil") ;
您可以不做循环而遵循此操作
Set<String> results1 = new HashSet<String>(allWords);
results1.removeAll(list1);
System.out.println("result1 : "+ list1 +" ===> "+ results1);
Set<String> results2 = new HashSet<String>(allWords);
results2.removeAll(list2);
System.out.println("result2 : "+ list2 +" ===> "+ results2);
Set<String> results3 = new HashSet<String>(allWords);
results3.removeAll(list3);
System.out.println("result3 : "+ list3 +" ===> "+ results3);
希望这有帮助我认为使用等于匹配字符串可以避免子字符串问题
for(String a : allWords)
{
for (int j = 0; j < subsets.size(); j++)
{
boolean isMatch = false;
String subSetString = "";
for(int k = 0 ; k < subsets.get(j).size() ; k++)
{
if(subsets.get(j).get(k).equals(a))
{
isMatch = true;
break;
}
subSetString += subsets.get(j).get(k) + " ";
}
if(!isMatch)
{
System.out.println(subSetString + " => " + a);
}
}
}
for(字符串a:allWords)
{
对于(int j=0;j”+a);
}
}
}
< /代码>发布一些可运行代码供我们调试。请显示您的代码,以便我们可以帮助您。也许您应该搜索“钢笔”的“钢笔”?考虑提供一个演示您的问题。这将减少混乱和更好的响应。是否使用字符串包含?为什么不简单地使用列表包含?
List<String> allWords = new ArrayList<>();
allWords.add("Book"); allWords.add("Pen"); allWords.add("Pencil") ;
list1 [Book,Pen]
list2 [Book,Pencil]
list3 [Pen,Pencil]
Set<String> results1 = new HashSet<String>(allWords);
results1.removeAll(list1);
System.out.println("result1 : "+ list1 +" ===> "+ results1);
Set<String> results2 = new HashSet<String>(allWords);
results2.removeAll(list2);
System.out.println("result2 : "+ list2 +" ===> "+ results2);
Set<String> results3 = new HashSet<String>(allWords);
results3.removeAll(list3);
System.out.println("result3 : "+ list3 +" ===> "+ results3);
result1 : [Book, Pen] ===> [Pencil]
result2 : [Book, Pencil] ===> [Pen]
result3 : [Pen, Pencil] ===> [Book]
for(String a : allWords)
{
for (int j = 0; j < subsets.size(); j++)
{
boolean isMatch = false;
String subSetString = "";
for(int k = 0 ; k < subsets.get(j).size() ; k++)
{
if(subsets.get(j).get(k).equals(a))
{
isMatch = true;
break;
}
subSetString += subsets.get(j).get(k) + " ";
}
if(!isMatch)
{
System.out.println(subSetString + " => " + a);
}
}
}