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用來(lái)自字符串的輸入填充二維數(shù)組的行

Java

慕運(yùn)維8079593 2022-07-27 11:23:41

我有以下問(wèn)題：給定矩陣的每一行的值，列由空格分隔 - 所以我在字符串?dāng)?shù)組中輸入所有行值，刪除空格并將數(shù)字解析為 int 數(shù)組?，F(xiàn)在每一行的值看起來(lái)像 1 個(gè)數(shù)字“12345”，而它們應(yīng)該是“1 2 3 4 5”。如何首先分隔數(shù)字，然后通過(guò)將元素添加到每一行來(lái)填充我的矩陣？謝謝！這是我的代碼： String n1 = input.nextLine (); int n = Integer.parseInt(n1); //rows of the matrix String[] arr = new String [n]; //contains all the rows of the matrix int [] array = new int [arr.length]; // contains all the elements of the rows of the matrix without whitespace for (int i = 0; i < arr.length; i++) { arr [i] = input.nextLine().replaceAll("\\s+",""); array[i] = Integer.parseInt(arr[i]); } int matrix [][] = new int [n][arr[0].length()];

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3 回答

holdtom

TA貢獻(xiàn)1805條經(jīng)驗(yàn) 獲得超10個(gè)贊

您應(yīng)該split()通過(guò)一些字符輸入字符串（在您的示例中為空格）。

示例如何轉(zhuǎn)換String為數(shù)組String（使用split()方法）

// Example input

String input = "1 2 3 4 5";

// Split elements by space

// So you receive array: {"1", "2", "3", "4", "5"}

String[] numbers = input.split(" ");

for (int position = 0; position < numbers.length; position++) {

// Get element from "position"

System.out.println(numbers[position]);

}

示例如何轉(zhuǎn)換String為數(shù)組int

// Example input

String input = "1 2 3 4 5";

// Split elements by space

// So you receive array: {"1", "2", "3", "4", "5"}

String[] strings = input.split(" ");

// Create new array for "ints" (with same size!)

int[] number = new int[strings.length];

// Convert all of the "Strings" to "ints"

for (int position = 0; position < strings.length; position++) {

number[position] = Integer.parseInt(strings[position]);

}

反對(duì) 回復(fù) 2022-07-27

侃侃無(wú)極

TA貢獻(xiàn)2051條經(jīng)驗(yàn) 獲得超10個(gè)贊

很難說(shuō)，但據(jù)我了解，您正試圖通過(guò)掃描儀逐行輸入矩陣。這可以解決你的問(wèn)題。

Scanner scanner = new Scanner(System.in);

//number of rows

int n = Integer.parseInt(scanner.nextLine());

int[][] matrix = new int[n][];

for(int i=0;i<n;i++) {

String line = scanner.nextLine();

String[] numbers = line.split(" ");

matrix[i] = new int[numbers.length];

for(int j=0;j<numbers.length;j++) {

matrix[i][j] = Integer.parseInt(numbers[j]);

}

反對(duì) 回復(fù) 2022-07-27

長(zhǎng)風(fēng)秋雁

TA貢獻(xiàn)1757條經(jīng)驗(yàn) 獲得超7個(gè)贊

在這里你有重要的問(wèn)題：

for (int i = 0; i < arr.length; i++) {

arr [i] = input.nextLine().replaceAll("\\s+",""); // loses the separator between the number

array[i] = Integer.parseInt(arr[i]); // makes no sense as you want get all numbers submitted for the current row and no a single one

}

如果您在提交的每一行填充矩陣，則可以使用更少的變量來(lái)進(jìn)行處理。

沒(méi)有經(jīng)過(guò)測(cè)試的代碼，但您應(yīng)該有所了解。

String n1 = input.nextLine();

int n = Integer.parseInt(n1); //rows of the matrix

int matrix [][] = null; // init it later : as you would have the two dimensions knowledge

for (int i = 0; i < n; i++) {

String[] numberToken = input.nextLine().split("\\s");

// matrix init : one time

if (matrix == null){ matrix [][] = new int[n][numberToken.length]; }

// array of int to contain numbers of the current row

int[] array = new int[numberToken.length];

// map String to int. Beware exception handling that you should do

for (int j = 0; j < numberToken.length; j++){

array[j] = Integer.parseInt(numberToken[j]);

}

// populate current row of the matrix

matrix[i] = array[j];

}

反對(duì) 回復(fù) 2022-07-27

3 回答
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用來(lái)自字符串的輸入填充二維數(shù)組的行

用來(lái)自字符串的輸入填充二維數(shù)組的行

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