英文:
How to check if a char is upper/lowercase?
问题
以下是翻译好的代码部分：
static int strScore(String str[], String s, int n) {
int score = 0, index = 0;
for (int i = 0; i < n; i++) {
if (str[i].equals(s)) {
for (int j = 0; j < s.length(); j++)
score += s.charAt(j) - 'a' + 1;
index = i + 1;
break;
}
}
score = score * index;
return score;
}
public static void main(String[] args) {
String str[] = { "abcde" };
String s = "abcde";
int n = str.length;
int score = strScore(str, s, n);
System.out.println(score);
}
英文:
The following code is supposed to convert letters to numbers and give the sum, but ignore any letters that are uppercase.
Example:
The input abcde should return 15. The input abCde should return 12.
Any help is appreciated.
static int strScore(String str[], String s, int n) {
int score = 0, index=0;
for (int i = 0; i < n; i++) {
if (str[i] == s) {
for (int j = 0; j < s.length(); j++)
score += s.charAt(j) - 'a' + 1;
index = i + 1;
break;
}
}
score = score * index;
return score;
}
public static void main(String[] args) {
String str[] = { "abcde" };
String s = "abcde";
int n = str.length;
int score = strScore(str, s, n);
System.out.println( score);
}
</details>
# 答案1
**得分**: 4
使用`Character.isLowerCase(...)`。
因此，您的`strScore`方法应该如下所示：
```java
static int strScore(String str[], String s, int n) {
int score = 0, index = 0;
for (int i = 0; i < n; i++) {
if (str[i].equals(s)) {
for (int j = 0; j < s.length(); j++) {
char c = s.charAt(j);
if(Character.isLowerCase(c)) // <-- 这部分很重要
score += c - 'a' + 1;
}
index = i + 1;
break;
}
}
score = score * index;
return score;
}
正如评论中指出的，不需要str参数，因此也不需要n参数。以下是更好的版本：
static int strScore(String s) {
int score = 0;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(Character.isLowerCase(c))
score += c - 'a' + 1;
}
return score;
}
英文:
Use Character.isLowerCase(...).
So this is what your strScore method should look like:
static int strScore(String str[], String s, int n) {
int score = 0, index = 0;
for (int i = 0; i < n; i++) {
if (str[i].equals(s)) {
for (int j = 0; j < s.length(); j++) {
char c = s.charAt(j);
if(Character.isLowerCase(c)) // <-- This is the important part
score += c - 'a' + 1;
}
index = i + 1;
break;
}
}
score = score * index;
return score;
}
As pointed out in the comments, there is no need for the str and therfore neither the n parameter. This is a better version:
static int strScore(String s) {
int score = 0;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(Character.isLowerCase(c))
score += c - 'a' + 1;
}
return score;
}
答案2
得分: 2
有两个问题需要解决：
- 您使用
==来比较字符串。您需要使用.equals。 - 您需要像这样进行检查
if(s.charAt(j)>= 'a' && s.charAt(j)<'z')
for (int i = 0; i < n; i++) {
if (str[i].equals(s)) {
for (int j = 0; j < s.length(); j++)
if(s.charAt(j)>= 'a' && s.charAt(j)<'z') {
score += s.charAt(j) - 'a' + 1;
英文:
There are two things to address:
- You have used
==to compare strings. You need to use.equals - You need to put a check like
if(s.charAt(j)>= 'a' && s.charAt(j)<'z')
for (int i = 0; i < n; i++) {
if (str[i].equals(s)) {
for (int j = 0; j < s.length(); j++)
if(s.charAt(j)>= 'a' && s.charAt(j)<'z') {
score += s.charAt(j) - 'a' + 1;
答案3
得分: 1
You can avoid passing String str[] = { "abcde" }; which has one element which equals s to The method. You can also avoid passing n which is an simply str.length():
static int strScore(String s) {
int score = 0, index = 0;
for (int i = 0; i < s.length(); i++) {
for (char c : s.toCharArray()) {
if(c >= 'a' && c < 'z') { //alternatively if(Character.isLowerCase(c))
score += c - 'a' + 1;
}
}
index = i + 1;
break;
}
score = score * index;
return score;
}
英文:
You can avoid passing String str[] = { "abcde" }; which has one element which equals s
to The method. You can also avoid passing n which is an simply str.length():
static int strScore(String s) {
int score = 0, index = 0;
for (int i = 0; i < s.length(); i++) {
for (char c : s.toCharArray()) {
if(c >= 'a' && c <'z') { //alternatively if(Character.isLowerCase(c))
score += c - 'a' + 1;
}
}
index = i + 1;
break;
}
score = score * index;
return score;
}
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