英文:
Java Looking for specific letters in a sting
问题
以下是您的代码的翻译部分：
我在想如何迭代遍历一个字符串并检查有多少个 "hi" 出现例如，如果字符串是 "hihi"，则计数应该输出 2。这是我目前的代码public static int countHi(String str) {int counter = 0;for (int i = 0; i < str.length(); i++) {if (str.substring(i, i + 1).equals("h")) {if (str.substring(i, i + 1).equals("i")) {counter = counter + 1;}}}return counter;}public static void main(String[] args) {String str = "hihi";int i = countHi(str);System.out.println("number of hi = " + i);}
英文:
I was wondering how to iterate over a string and to check how many hi's come out
For example, if the string is "hihi" the count should output 2.
This is what I have so far
public static int countHi(String str) {int counter = 0;for (int i = 0; i < str.length(); i++) {if (str.substring(i, i + 1) == "h") {if (str.substring(i, i + 1) == "i") {counter = counter + 1;}}}return counter;}public static void main(String[] args) {String str = "hihi";int i = countHi(str);System.out.println("number of hi = " + i);}
答案1
得分: 1
你应该使用.equals来比较实例（例如 String），而不是使用==。然而，在这里，你可以在String.charAt(int)中使用==。另外，我建议从第二个字符开始，将当前索引处的字符与i以及前一个索引处的字符与h进行比较。像这样：
public static int countHi(String str) {int counter = 0;for (int i = 1; i < str.length(); i++) {if (str.charAt(i - 1) == 'h' && str.charAt(i) == 'i') {counter++;}}return counter;}
或者，你可以将当前索引处的字符与h进行比较，将下一个索引处的字符与i进行比较（但现在需要提前停止迭代一个字符）。像这样：
public static int countHi(String str) {int counter = 0;for (int i = 0; i < str.length() - 1; i++) {if (str.charAt(i) == 'h' && str.charAt(i + 1) == 'i') {counter++;}}return counter;}
英文:
You compare instances (like String) with .equals (not ==). However, here you can use == with String.charAt(int). Also, I would start with the second character and compare the character at the current index with i and the previous index with h. Like,
public static int countHi(String str) {int counter = 0;for (int i = 1; i < str.length(); i++) {if (str.charAt(i - 1) == 'h' && str.charAt(i) == 'i') {counter++;}}return counter;}
Alternatively, compare the character at the current index with h and the character at the next index with i (but now you need to stop iterating a character earlier). Like,
public static int countHi(String str) {int counter = 0;for (int i = 0; i < str.length() - 1; i++) {if (str.charAt(i) == 'h' && str.charAt(i + 1) == 'i') {counter++;}}return counter;}
答案2
得分: 0
这是简单的方法：
public static int countHi(String str) {return split(str, -1).length - 1;}
请注意，您必须将split()的第二个参数传递为-1；如果没有这个参数，结果中的尾随空格将被删除。
英文:
Here's the easy way:
public static int countHi(String str) {return split(str, -1).length - 1;}
Note that you must pass -1 as the second parameter of split(); without it, trailing blanks would be pruned from the result.
答案3
得分: 0
为了紧凑和可读性，也许可以：
int count = 0;Matcher matcher = Pattern.compile("hi").matcher(string)while (matcher.find()) {count++;}
这种方法适用于任何正则表达式模式，尽管不是最高效的。
英文:
For compactness and readability, maybe:
int count = 0;Matcher matcher = Pattern.compile(“hi”).matcher(string)while (matcher.find()) {count++;}
This approach will work for any Regular Expression pattern, although it won’t be the most efficient.
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