问题
The following code will print 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"
The expected result would be: [2,3,5]
回答1:
This should print the list of positions without the -1 at the end that Peter Lawrey's solution has had.
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);
}
[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.]
回答2:
Try the following (Which does not print -1 at the end now!)
int index = word.indexOf(guess);
while(index >= 0) {
System.out.println(index);
index = word.indexOf(guess, index+1);
}
回答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 :
for(Integer i : list){
System.out.println(i);
}
Or as a array :
list.toArray();
回答4:
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);
回答5:
int index = -1;
while((index = text.indexOf("on", index + 1)) >= 0) {
LOG.d("index=" + index);
}
回答6:
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);
}
回答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);
回答8:
Also, if u want to find all indexes of a String in a String.
int index = word.indexOf(guess);
while (index >= 0) {
System.out.println(index);
index = word.indexOf(guess, index + guess.length());
}
回答9:
I had this problem as well, until I came up with this method.
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:
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);
}
}
回答10:
A class for splitting strings I came up with. A short test is provided at the end.
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.
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;
}
}
Simple test code:
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();
}
回答11:
This can be done by iterating myString and shifting fromIndex parameter in indexOf():
int currentIndex = 0;
while (
myString.indexOf(
mySubstring,
currentIndex) >= 0) {
System.out.println(currentIndex);
currentIndex++;
}
回答12:
Try this
String str = "helloslkhellodjladfjhello";
String findStr = "hello";
System.out.println(StringUtils.countMatches(str, findStr));
来源：https://stackoverflow.com/questions/5034442/indexes-of-all-occurrences-of-character-in-a-string