所以我在java編程課中學(xué)習(xí)數(shù)組，并且我得到了一個(gè)對(duì)我來說非常具有挑戰(zhàn)性的程序。我必須編寫一個(gè)程序，其中包含一系列行的輸入文件，每行一個(gè)單詞。我必須編寫一個(gè)程序，告訴用戶該單詞是否使用不同的字母（重復(fù)的字母）。這是我的輸入文件：UNCOPYRIGHTABLE
FLIPPER
EXECUTABLE
UNPROFITABLE
QUESTIONABLE
WINDOW
TAMBOURINE這就是我現(xiàn)在所擁有的：Scanner df = new Scanner (new Files (distinctlet.in"));while (df.hasNextLine()){
String line = df.nextLine();
String array [] = line.split("");
String ans = "";for (int k = 0; k < array.length; k++){
for (int m = k + 1; m < array.length; m++)
{
if (!array[k].equals(array[m])
{
ans = "USES DISTINCT LETTERS";
}
else
{
ans = "DOES NOT USE DISTINCT LETTERS";
}
}//FOR LOOP2}//FOR LOOPSystem.out.println(line + " " + ans);}//WHILE DF我的輸出應(yīng)該是：UNCOPYRIGHTABLE USES DISTINCT LETTERS
FLIPPER DOES NOT USE DISTINCT LETTERS
EXECUTABLE DOES NOT USE DISTINCT LETTERS等等...我現(xiàn)在的輸出是每一行的輸入和“不要使用獨(dú)特的字母”。我知道問題是在嵌套循環(huán)中，但我不知道如何解決這個(gè)問題。感謝您的幫助。
5 回答
慕勒3428872
TA貢獻(xiàn)1848條經(jīng)驗(yàn) 獲得超6個(gè)贊
您可以使用字符位置來確定它是否存在于字符串中的其他位置。考慮以下解決方案：
while (df.hasNextLine()) {
String line = df.nextLine();
String array[] = line.split("");
String ans = "USES DISTINCT LETTERS";
for (int k = 0; k < array.length; k++) {
if(line.indexOf(array[k]) != line.lastIndexOf(array[k])){
ans = "DOES NOT USE DISTINCT LETTERS";
break;
}
}//FOR LOOP
System.out.println(line + " " + ans);
}//WHILE DF
婷婷同學(xué)_
TA貢獻(xiàn)1844條經(jīng)驗(yàn) 獲得超8個(gè)贊
要解決這類問題，這個(gè)解決方案可以很容易地解決你的問題，但這里我們只是采用一個(gè)256長度數(shù)組的常量空間complexity will be O(n)：
int []star = new int[256];
while (df.hasNextLine())
{
Arrays.fill(star,0);
String line = df.nextLine();
for(int i=0;i<line.length();i++){
star[line.charAt(0)]++;
}
for(int i=0;i<256;i++){
if(star[i]>0 && star[i]>1){
System.out.println("Duplicate characters present..");
}
}
System.out.println("No Duplicate characters present..");
}我希望你有個(gè)主意..
守著星空守著你
TA貢獻(xiàn)1799條經(jīng)驗(yàn) 獲得超8個(gè)贊
就個(gè)人而言，我不會(huì)使用數(shù)組來做到這一點(diǎn) - 我會(huì)使用地圖。即使你必須使用數(shù)組，我仍然會(huì)以地圖的精神解決這個(gè)問題。
這里的counter數(shù)組就像一個(gè)map（key = index，value = count）。
public class Test {
public static void main(String[] args) throws IOException {
byte[] encoded = Files.readAllBytes(Paths.get("data/input.csv"));
String s = new String(encoded, Charset.defaultCharset());
String[] split = s.split("\n");
System.out.println("Input: " + Arrays.toString(split));
System.out.println("Output: " + Arrays.toString(check(split)));
}
private static String[] check(String[] strings) {
for (int i = 0; i < strings.length; i++)
strings[i] += distinct(strings[i])
? " USES DISTINCT LETTERS"
: " DOES NOT USE DISTINCT LETTERS";
return strings;
}
private static boolean distinct(String string) {
int[] counter = new int[string.length()];
for (char c : string.toCharArray())
if (++counter[string.indexOf(c)] > 1) return false;
return true;
}}Input: [UNCOPYRIGHTABLE, FLIPPER, EXECUTABLE, UNPROFITABLE, QUESTIONABLE, WINDOW, TAMBOURINE]
Output: [UNCOPYRIGHTABLE USES DISTINCT LETTERS, FLIPPER DOES NOT USE DISTINCT LETTERS, EXECUTABLE DOES NOT USE DISTINCT LETTERS, UNPROFITABLE USES DISTINCT LETTERS, QUESTIONABLE DOES NOT USE DISTINCT LETTERS, WINDOW DOES NOT USE DISTINCT LETTERS, TAMBOURINE USES DISTINCT LETTERS]
慕姐4208626
TA貢獻(xiàn)1852條經(jīng)驗(yàn) 獲得超7個(gè)贊
for (int k = 0; k < array.length; k++)
{
for (int m = k + 1; m < array.length; m++)
{
if (!array[k].equals(array[m])
{
ans = "USES DISTINCT LETTERS";
}
else
{
ans = "DOES NOT USE DISTINCT LETTERS";
// Break out of the two loops once you know the characters are matching.
//Otherwise it loops again and the last match of character is what you get the ans.
}
}//FOR LOOP2
}//FOR LOOP
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