字符串中所有出现的字符的索引

下面的代码将打印2

String word = "bannanas";
String guess = "n";
int index;
System.out.println(
index = word.indexOf(guess)
);

我想知道如何获取字符串“ bananas”中的“ n”(猜测)的所有索引

预期的结果是: [2,3,5]

220835 次浏览
小开 
String string = "bannanas";
ArrayList<Integer> list = new ArrayList<Integer>();
char character = 'n';
for(int i = 0; i < string.length(); i++){
if(string.charAt(i) == character){
list.add(i);
}
}

结果应该是这样的:

    for(Integer i : list){
System.out.println(i);
}

或者作为一个数组:

list.toArray();
小开

尝试以下操作(现在不会在结尾打印 -1!)

int index = word.indexOf(guess);
while(index >= 0) {
System.out.println(index);
index = word.indexOf(guess, index+1);
}
小开 
String word = "bannanas";


String guess = "n";


String temp = word;


while(temp.indexOf(guess) != -1) {
int index = temp.indexOf(guess);
System.out.println(index);
temp = temp.substring(index + 1);
}
小开
最佳答案

这应该打印的位置清单没有 -1在结束时，Peter Lawrey 的解决方案 已经有。

int index = word.indexOf(guess);
while (index >= 0) {
System.out.println(index);
index = word.indexOf(guess, index + 1);
}

It can also be done as a for loop:

for (int index = word.indexOf(guess);
index >= 0;
index = word.indexOf(guess, index + 1))
{
System.out.println(index);
}

[注意: 如果 guess可以比单个字符长，那么通过分析 guess字符串，可以比上面的循环更快地遍历 word。这种方法的基准是 Boyer-Moore字符串搜索算法。然而，有利于使用这种方法的条件似乎并不存在。]

小开 
    String input = "GATATATGCG";
String substring = "G";
String temp = input;
String indexOF ="";
int tempIntex=1;


while(temp.indexOf(substring) != -1)
{
int index = temp.indexOf(substring);
indexOF +=(index+tempIntex)+" ";
tempIntex+=(index+1);
temp = temp.substring(index + 1);
}
Log.e("indexOf ","" + indexOF);
小开 
int index = -1;
while((index = text.indexOf("on", index + 1)) >= 0) {
LOG.d("index=" + index);
}
小开

试试这个

String str = "helloslkhellodjladfjhello";
String findStr = "hello";


System.out.println(StringUtils.countMatches(str, findStr));
小开

另外，如果您想在 String 中查找 String 的所有索引。

int index = word.indexOf(guess);
while (index >= 0) {
System.out.println(index);
index = word.indexOf(guess, index + guess.length());
}
小开

这可以通过在 indexOf()中迭代 myString并移动 fromIndex参数来实现:

  int currentIndex = 0;


while (
myString.indexOf(
mySubstring,
currentIndex) >= 0) {


System.out.println(currentIndex);


currentIndex++;
}
小开

我也有这个问题，直到我想出了这个方法。

public static int[] indexesOf(String s, String flag) {
int flagLen = flag.length();
String current = s;
int[] res = new int[s.length()];
int count = 0;
int base = 0;
while(current.contains(flag)) {
int index = current.indexOf(flag);
res[count] = index + base;
base += index + flagLen;
current = current.substring(current.indexOf(flag) + flagLen, current.length());
++ count;
}
return Arrays.copyOf(res, count);
}

此方法可用于查找字符串中任意长度的任何标志的索引，例如:

public class Main {


public static void main(String[] args) {
int[] indexes = indexesOf("Hello, yellow jello", "ll");


// Prints [2, 9, 16]
System.out.println(Arrays.toString(indexes));
}


public static int[] indexesOf(String s, String flag) {
int flagLen = flag.length();
String current = s;
int[] res = new int[s.length()];
int count = 0;
int base = 0;
while(current.contains(flag)) {
int index = current.indexOf(flag);
res[count] = index + base;
base += index + flagLen;
current = current.substring(current.indexOf(flag) + flagLen, current.length());
++ count;
}
return Arrays.copyOf(res, count);
}
}
小开

我想出的一个分解字符串的类。最后提供了一个简短的测试。

如果可能，SplitStringUtils.smartSplitToShorterStrings(String str, int maxLen, int maxParts)将按空格分割，而不会中断单词; 如果不能，则按照 maxLen 按索引分割。

Other methods provided to control how it is split: bruteSplitLimit(String str, int maxLen, int maxParts), spaceSplit(String str, int maxLen, int maxParts).

public class SplitStringUtils {


public static String[] smartSplitToShorterStrings(String str, int maxLen, int maxParts) {
if (str.length() <= maxLen) {
return new String[] {str};
}
if (str.length() > maxLen*maxParts) {
return bruteSplitLimit(str, maxLen, maxParts);
}


String[] res = spaceSplit(str, maxLen, maxParts);
if (res != null) {
return res;
}


return bruteSplitLimit(str, maxLen, maxParts);
}


public static String[] bruteSplitLimit(String str, int maxLen, int maxParts) {
String[] bruteArr = bruteSplit(str, maxLen);
String[] ret = Arrays.stream(bruteArr)
.limit(maxParts)
.collect(Collectors.toList())
.toArray(new String[maxParts]);
return ret;
}


public static String[] bruteSplit(String name, int maxLen) {
List<String> res = new ArrayList<>();
int start =0;
int end = maxLen;
while (end <= name.length()) {
String substr = name.substring(start, end);
res.add(substr);
start = end;
end +=maxLen;
}
String substr = name.substring(start, name.length());
res.add(substr);
return res.toArray(new String[res.size()]);
}


public static String[] spaceSplit(String str, int maxLen, int maxParts) {
List<Integer> spaceIndexes = findSplitPoints(str, ' ');
List<Integer> goodSplitIndexes = new ArrayList<>();
int goodIndex = -1;
int curPartMax = maxLen;
for (int i=0; i< spaceIndexes.size(); i++) {
int idx = spaceIndexes.get(i);
if (idx < curPartMax) {
goodIndex = idx;
} else {
goodSplitIndexes.add(goodIndex+1);
curPartMax = goodIndex+1+maxLen;
}
}
if (goodSplitIndexes.get(goodSplitIndexes.size()-1) != str.length()) {
goodSplitIndexes.add(str.length());
}
if (goodSplitIndexes.size()<=maxParts) {
List<String> res = new ArrayList<>();
int start = 0;
for (int i=0; i<goodSplitIndexes.size(); i++) {
int end = goodSplitIndexes.get(i);
if (end-start > maxLen) {
return null;
}
res.add(str.substring(start, end));
start = end;
}
return res.toArray(new String[res.size()]);
}
return null;
}




private static List<Integer> findSplitPoints(String str, char c) {
List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == c) {
list.add(i);
}
}
list.add(str.length());
return list;
}
}

简单的测试代码:

  public static void main(String[] args) {
String [] testStrings = {
"123",
"123 123 123 1123 123 123 123 123 123 123",
"123 54123 5123 513 54w567 3567 e56 73w45 63 567356 735687 4678 4678 u4678 u4678 56rt64w5 6546345",
"1345678934576235784620957029356723578946",
"12764444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444",
"3463356 35673567567 3567 35 3567 35 675 653 673567 777777777777777777777777777777777777777777777777777777777777777777"
};


int max = 35;
int maxparts = 2;




for (String str : testStrings) {
System.out.println("TEST\n    |"+str+"|");
printSplitDetails(max, maxparts);
String[] res = smartSplitToShorterStrings(str, max, maxparts);
for (int i=0; i< res.length;i++) {
System.out.println("  "+i+": "+res[i]);
}
System.out.println("===========================================================================================================================================================");
}


}


static void printSplitDetails(int max, int maxparts) {
System.out.print("  X: ");
for (int i=0; i<max*maxparts; i++) {
if (i%max == 0) {
System.out.print("|");
} else {
System.out.print("-");
}
}
System.out.println();
}
小开

With Java9, one can make use of the iterate(int seed, IntPredicate hasNext,IntUnaryOperator next) as follows:-

List<Integer> indexes = IntStream
.iterate(word.indexOf(c), index -> index >= 0, index -> word.indexOf(c, index + 1))
.boxed()
.collect(Collectors.toList());
System.out.printlnt(indexes);
小开

这是一个 java 8解决方案。

public int[] solution (String s, String subString){
int initialIndex = s.indexOf(subString);
List<Integer> indexList = new ArrayList<>();
while (initialIndex >=0){
indexList.add(initialIndex);
initialIndex = s.indexOf(subString, initialIndex+1);
}
int [] intA = indexList.stream().mapToInt(i->i).toArray();
return intA;
}
小开

这可以通过使用正则表达式在 Java9中以函数的方式实现:

Pattern.compile(Pattern.quote(guess)) // sanitize input and create pattern
.matcher(word) // create matcher
.results()     // get the MatchResults, Java 9 method
.map(MatchResult::start) // get the first index
.collect(Collectors.toList()) // collect found indices into a list
);

Here's the Kotlin Solution to add this logic as a new a new methods into CharSequence API using extension method:

 // Extension method
fun CharSequence.indicesOf(input: String): List<Int> =
Regex(Pattern.quote(input)) // build regex
.findAll(this)          // get the matches
.map { it.range.first } // get the index
.toCollection(mutableListOf()) // collect the result as list


// call the methods as
"Banana".indicesOf("a") // [1, 3, 5]
小开

Java8 +

要在 String中查找特定字符的所有索引，可以创建所有索引的 IntStream，并在其上创建 filter。

import java.util.stream.Collectors;
import java.util.stream.IntStream;
//...
String word = "bannanas";
char search = 'n';
//To get List of indexes:
List<Integer> indexes = IntStream.range(0, word.length())
.filter(i -> word.charAt(i) == search).boxed()
.collect(Collectors.toList());
//To get array of indexes:
int[] indexes = IntStream.range(0, word.length())
.filter(i -> word.charAt(i) == search).toArray();

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