问题描述
我正在解决一个简单的问题:
I was solving an easy question :
在Java中删除字符数组的某些字符，想法很简单:
Remove certain characters of a character array in Java , the idea is straightforward :
static void remove_char(char[] arr, char c){
int r = 0;
for (int i = 0 ; i < arr.length ; i++){
if (arr[i] == c){
r++;
continue;
}
arr[i - r] = arr[i];
}
arr[arr.length - r] = ' '; // ??
return;
}
我想放置一个 结束字符，它表示当我们想要生成字符串时不必考虑数组的其余部分，例如，使用 new String(arr)
I want to put an ending character which signals that the rest of the array doesn't have to be considered when we want to, for example, generate a string using new String(arr)
Java 中有这样的字符吗?(我猜在 C 中是 但我不确定)
Is there any such character in Java ? ( I guess in C it's but I am not sure)
例如当我们调用:
System.out.println(new String(remove_char(new char[] {'s','a','l','a','m'} ,'a')))
这将被打印出来:slm m
虽然我想获得 slm 并且我想 in-place 不使用新数组
While I want to get slm and I want to do this in-place not using a new array
推荐答案
Java 不像 C 那样标记"字符串的结尾.它跟踪长度和值，因此字符串中可能包含零字符 ( ).如果您从包含 字符的 char 数组创建 String，则生成的 String 将包含这些字符.
Java doesn't "mark" the end-of-string as C does. It tracks length & values, so it's possible to have zero-chars ( ) in the string. If you create a String from a char array containing chars, the resultant String will contain those characters.
另请注意，数组具有静态大小 - 它不能重新调整大小，因此您必须自己跟踪大小"(String 构造函数不会从 char 末尾为您删除不需要的字符数组).
Note also, an array has a static size - it cannot be re-sized, so you'll have to track the "size" yourself (the String constructor won't drop undesirable characters for you from the end of a char array).
考虑使用任一:
Clyde Byrd III 建议的
StringBuilder，或
String(char value[], int offset, int count);您必须以编程方式确定 offset 和 count 的适当值.也许字符串方法，例如 replace、concat 或 substring 可能会有所帮助.
String(char value[], int offset, int count); you'll have to determine programmatically what the appropriate values for offset and count would be. Perhaps String methods, such as replace, concat, or substring might help.
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