[英]I get java.lang.StringIndexOutOfBoundsException when translating c++ code into java code
我正在努力解決這個問題
https://vjudge.net/problem/UVALive-6805
我找到了解決方案,但在 c++ 中,誰能幫我將其轉換為 java 代碼。 我是編程的新手
我嘗試了很多解決方案,但沒有一個有效。
如果可能的話,我需要這方面的幫助
例如,我不知道 java 中 c++ 中.erase函數的等價物是什么? c++ 中的 sbstr 是否提供與 java 不同的結果?
#include <iostream>
#include <string>
using namespace std;
int syllable(string word)
{
int L = word.size();
int syllable;
if (L>=7)
{
syllable = 3;
}
else if (L==6)
{
int indicator = 0;
for (int k=0; k<=L-2; k++)
{
string subword = word.substr(k, 2);
if (subword == "ng" || subword == "ny")
{
indicator++;
}
}
if (indicator == 0)
{
syllable = 3;
}
else
{
syllable = 2;
}
}
else if (L == 4 || L == 5)
{
syllable = 2;
}
else if (L == 3)
{
char Char = word[0];
if (Char=='a' || Char=='A' || Char=='e' || Char=='E' || Char=='i' || Char=='I' || Char=='o' || Char=='O' || Char=='u' || Char=='U')
{
syllable = 2;
}
else
{
syllable = 1;
}
}
else
{
syllable = 1;
}
return syllable;
}
int main()
{
string word;
int T;
cin >> T;
for (int i=1; i<=T; i++)
{
int syl[] = {0, -1, -2, -3};
string rhy[] = {"a", "b", "c", "d"};
int verse = 0;
int stop = 0;
while (stop == 0)
{
cin >> word;
int L = word.size();
char end = word[L-1];
if (end == '.')
{
stop = 1;
}
if (word[L-1] == ',' || word[L-1] == '.')
{
word = word.erase(L-1, 1);
L = word.size();
}
if (verse<=3)
{
syl[verse] = syl[verse] + syllable(word);
}
if (end == ',' || end == '.')
{
if (verse<=3)
{
rhy[verse] = word.substr(L-2, 2);
}
verse++;
if (verse<=3)
{
syl[verse] = 0;
}
}
}
int A = 0, B = 0, C = 0, D = 0;
for (int k=0; k<4; k++)
{
if (syl[k] >= 8 && syl[k] <= 12)
{
A = A + 10;
}
}
for (int k=0; k<2; k++)
{
if (rhy[k] == rhy[k+2])
{
B = B + 20;
}
}
for (int k=0; k<2; k++)
{
if (syl[k] == syl[k+2])
{
C = C + 10;
}
}
if (verse > 4)
{
D = (verse - 4) * 10;
}
int E = A + B + C - D;
cout << "Case #" << i << ": " << A << " " << B << " " << C << " " << D << " " << E << endl;
}
}
這是我的嘗試
import java.util.*;
public class First {
public static int syllable(String word) {
int L = word.length();
int syllable;
if (L >= 7) {
syllable = 3;
} else if (L == 6) {
int indicator = 0;
for (int k = 0; k < L - 3; k++) {
String subword = word.substring(k, 2);
if (subword == "ng" || subword == "ny") {
indicator++;
}
}
if (indicator == 0) {
syllable = 3;
} else {
syllable = 2;
}
} else if (L == 4 || L == 5) {
syllable = 2;
} else if (L == 3) {
char Char = word.charAt(0);
if (Char == 'a' || Char == 'A' || Char == 'e' || Char == 'E' || Char == 'i' || Char == 'I' || Char == 'o'
|| Char == 'O' || Char == 'u' || Char == 'U') {
syllable = 2;
} else {
syllable = 1;
}
} else {
syllable = 1;
}
return syllable;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String word;
int T;
T = sc.nextInt();
for (int i = 1; i <= T; i++) {
int syl[] = { 0, -1, -2, -3 };
String rhy[] = { "a", "b", "c", "d" };
int verse = 0;
int stop = 0;
while (stop == 0) {
word = sc.next();
int L = word.length();
char end = word.charAt(L-1);
if (end == '.') {
stop = 1;
}
if (word.charAt(L-1) == ',' || word.charAt(L-1) == '.') {
word.substring(L-1, 1);
L = word.length();
}
if (verse <= 3) {
syl[verse] = syl[verse] + syllable(word);
}
if (end == ',' || end == '.') {
if (verse <= 3) {
rhy[verse] = word.substring(L - 2, 2);
}
verse++;
if (verse <= 3) {
syl[verse] = 0;
}
}
}
int A = 0, B = 0, C = 0, D = 0;
for (int k = 0; k < 4; k++) {
if (syl[k] >= 8 && syl[k] <= 12) {
A = A + 10;
}
}
for (int k = 0; k < 2; k++) {
if (rhy[k] == rhy[k + 2]) {
B = B + 20;
}
}
for (int k = 0; k < 2; k++) {
if (syl[k] == syl[k + 2]) {
C = C + 10;
}
}
if (verse > 4) {
D = (verse - 4) * 10;
}
int E = A + B + C - D;
System.out.println("Case #" + i + ": " + A + " " + B + " " + C + " " + D + " " + E);
}
}
}
異常由您第二次和第三次調用 String substring 方法引發。 您的 beginIndex 高於您的 endIndex。 正如你在這里看到的https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int,%20int) beginIndex 總是必須低於 endIndex。
在回答您的問題之前,一般來說,關於字符串和 Java,有一些要點需要提及。
java中的substring方法有兩種形式。
beginIndex
並返回從beginIndex
到str.length() - 1
(其中str
代表一個字符串)beginIndex
和endIndex
,並返回從beginIndex
到endIndex - 1
。 beginIndex
永遠不應該大於endIndex
否則它會拋出一個IndexOutOfBoundsException C++ 的 substring 方法 ( string::substr()
) 接受開頭的“索引”並接受它后面的字符數以包含在子字符串中。 因此,通過執行substr(L-2, 2)
可以獲得字符串的最后兩個字符。
Java 永遠不會讓你越界。 這意味着您需要不斷檢查您是否在您正在迭代的任何內容的范圍內。
考慮到所有這些,我會去驗證所有substring()
方法調用都返回了正確的字符范圍,並且您正確地將substring()
返回的值重新分配給了正確的變量。
為了模仿 C++ 的string::erase()
,根據您要擦除的單詞的哪個部分,您需要獲取之前部分的子字符串和之后部分的子字符串,並將它們加在一起。
前任。 假設我有String line = "I do not like the movies";
因為它是任何人都不可能不喜歡看電影,我們要切出的字not
我們通過做我上面說的來做到這一點
String before = line.substring(0, 5); // This gives us "I do " since it goes up to but not including the 5th index.
String after = line.substring(5 + 3); // This gives us the rest of the string starting after the word "not" because not is 3 characters long and this skips to the 3rd index after index 5 (or index 8)
line = before + after; // This'll add those two Strings together and give you "I do like the movies"
希望這可以幫助!
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