在Java实践中修改字符串
嗨,我在上面的代码上遇到了问题,它;这是一个实践问题,但“将直线从空间变为结尾的子字符串”确实让我困惑。我知道如何找到空间的索引,并从开始到索引打印字符串 预期产出:在Java实践中修改字符串,java,string,methods,Java,String,Methods,嗨,我在上面的代码上遇到了问题,它;这是一个实践问题,但“将直线从空间变为结尾的子字符串”确实让我困惑。我知道如何找到空间的索引,并从开始到索引打印字符串 预期产出: public static void displayWords(String line) { // while line has a length greater than 0 // find the index of the space. // if there is one, // print t
public static void displayWords(String line)
{
// while line has a length greater than 0
// find the index of the space.
// if there is one,
// print the substring from 0 to the space
// change line to be the substring from the space to the end
// else
// print line and then set it equal to the empty string
}
line = "I love you";
System.out.printf("\n\n\"%s\" contains the words:\n", line );
displayWords(line);
另一种方法
I
love
you
<将其用作参考。您需要使用substring()
方法。检查:
公共字符串子字符串(int-beginIndex,int-endIndex)
返回作为此字符串的子字符串的新字符串。子字符串从指定的beginIndex
开始,并延伸到索引endIndex-1
处的字符。因此,子字符串的长度是endIndex beginIndex
示例:
返回“汉堡包”。子字符串(4,8)
“敦促”
返回“微笑”。子字符串(1,5)
“英里”
public static void displayWords(String line)
{
Scanner s = new Scanner(line)
while (s.hasNext())
{
System.out.println(s.next())
}
}
你有足够的时间。。。今晚我感到很慷慨。因此,解决方案如下:
public static void displayWords(String line) {
while(line.length() > 0) { // This is the first part (the obvious one)
/*
You should do something here that tracks the index of the first space
character in the line. I suggest you check the `charAt()` method, and use
a for() loop.
Let's say you find a space at index i
*/
System.out.println(line.substring(0,i)); // This prints the substring of line
// that goes from the beginning
// (index 0) to the position right
// before the space
/*
Finally, here you should assign a new value to line, that value must start
after the recently found space and end at the end of the line.
Hint: use substring() again, and remember that line.lengh() will give you
the position of the last character of line plus one.
*/
}
}
publicstaticvoiddisplaywords(字符串行){
int i;//您需要这个
while(line.length()>0){//这是第一部分(显而易见的一部分)
/*
在这里,检查每个字符,如果找到空格,则中断循环
*/
对于(i=0;i
在再次阅读您的问题后,我意识到它应该是一个递归解决方案。。。下面是递归方法:
public static void displayWords(String line) {
int i; // You'll need this
while(line.length() > 0) { // This is the first part (the obvious one)
/*
Here, check each character and if you find a space, break the loop
*/
for(i = 0; i < line.length(); i++) {
if(line.charAt(i) == ' ') // If the character at index i is a space ...
break; // ... break the for loop
}
System.out.println(line.substring(0,i)); // This prints the substring of line
// that goes from the beginning
// (index 0) to the position right
// before the space
/*
Here, the new value of line is the substring from the position of the space
plus one to the end of the line. If i is greater than the length of the line
then you're done.
*/
if(i < line.length())
line = line.substring(i + 1, line.length());
else
line = "";
}
}
publicstaticvoiddisplaywords(字符串行){
int i;
对于(i=0;i0)
显示字(行);
}
您是否查看了substring()
方法?我无法添加其他方法,这是一个实践问题,您必须使用提供给您的代码段。他的意思是我必须能够使用任何其他字符串,如“John God the store”,预期输出为“John God the store”。我可以使用substring方法打印单词,但它不适用于其他字符串。您可以通过扫描仪运行字符串。我认为OP是一个初学者,虽然这是一种非常简单的方法来完成所需的操作,但并不完全符合OP的需要。有时,在学习时,这样做是一件好事正确欣赏“简单、高效、干净、优雅的方式”的“艰难、不太有效的方式”
public static void displayWords(String line) {
int i;
for(i = 0; i < line.lengh(); i++) {
if(line.charAt(i) = ' ')
break;
}
System.out.println(line.substring(0, i);
if(i < line.lengh())
line = line.substring(i + 1, line.lengh();
else
line = "";
if(line.lengh() > 0)
displayWords(line);
}