字符串中所有出现的字符的索引

The following code will print 2

以下代码将打印2

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

I would like to know how to get all the indexes of "n" ("guess") in the string "bannanas"

我想知道如何在字符串“bannanas”中获取“n”（“guess”）的所有索引

The expected result would be: [2,3,5]

预期结果将是：[2,3,5]

12 个解决方案

#1

This should print the list of positions without the -1 at the end that Peter Lawrey's solution has had.

这应该打印出Peter Lierey的解决方案最终没有-1的位置列表。

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:

它也可以作为for循环完成：

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

[Note: if guess can be longer than a single character, then it is possible, by analyzing the guess string, to loop through word faster than the above loops do. The benchmark for such an approach is the Boyer-Moore algorithm. However, the conditions that would favor using such an approach do not seem to be present.]

[注意：如果猜测可能比单个字符长，那么通过分析猜测字符串，可以比上面的循环更快地循环通过单词。这种方法的基准是Boyer-Moore算法。但是，有利于使用这种方法的条件似乎并不存在。

#2

Try the following (Which does not print -1 at the end now!)

尝试以下（最后不打印-1！）

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

#3

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);
    }
}

Result would be used like this :

结果将如下使用：

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

Or as a array :

或者作为一个数组：

list.toArray();

#4

int index = -1;
while((index = text.indexOf("on", index + 1)) >= 0) {
   LOG.d("index=" + index);
}

#5

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);
}

#6

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

使用Java9，可以使用iterate（int seed，IntPredicate hasNext，IntUnaryOperator next），如下所示：

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);

#7

    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);

#8

Also, if u want to find all indexes of a String in a String.

此外，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());
}

#9

I had this problem as well, until I came up with this method.

我也有这个问题，直到我想出这个方法。

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);
}

This method can be used to find indexes of any flag of any length in a string, for example:

此方法可用于查找字符串中任何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);
    }
}

#10

A class for splitting strings I came up with. A short test is provided at the end.

我想出了一个用于分割字符串的类。最后提供了一个简短的测试。

SplitStringUtils.smartSplitToShorterStrings(String str, int maxLen, int maxParts) will split by spaces without breaking words, if possible, and if not, will split by indexes according to maxLen.

如果可能，SplitStringUtils.smartSplitToShorterStrings（String str，int maxLen，int maxParts）将按空格分割而不会破坏单词，否则，将根据maxLen按索引拆分。

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

提供其他方法来控制它的拆分方式：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;
  }
}

Simple test code:

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();
  }

#11

This can be done by iterating myString and shifting fromIndex parameter in indexOf():

这可以通过迭代myString并在indexOf（）中移动fromIndex参数来完成：

  int currentIndex = 0;

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

    System.out.println(currentIndex);

    currentIndex++;
  }

#12

Try this

尝试这个

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

Sy.