Java扫描器。NextInt（）跳过输入

Question

这里是Java新手，尝试扫描下面表单中的输入：

3

3 3
101
000
101

public static void main(String[] args) {

    Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
    int t = in.nextInt(); // t: 3, in.nextInt(): 3 as expected

    for (int i = 1; i <= t; i++) { // i: 1, t: 3, in.nextInt(): 3 (not sure which "3")
        int row = in.nextInt();
        // after this line in.nextInt(): 101 and row: 0 which are all WEIRD
        int column = in.nextInt();
        // now here it's in.nextInt(): 2 and column: 2

        int world[][] = new int[row][column];

        // in.nextInt(): 11 now and as we see it changed its value when I didn't even call it

        TreeMap<Integer, ArrayList<int[]>> distancesRandAcc = new TreeMap<Integer, ArrayList<int[]>>();
        distancesRandAcc.put(0, new ArrayList<int[]>());

        for (int j = 0; j < row; j++) {
            int temp = in.nextInt();
            char[] s = Integer.toString(temp).toCharArray();
            for (int k = 0; k < column; k++){
                world[j][k] = Character.getNumericValue(s[k]);
            }
        }
        int result = calculateDeliveryTime(world, row, column, distancesRandAcc);
        System.out.println("Case #" + i + ": " + result);
    }
}

拜托，有更有经验的人能教我如何解决这个问题吗？我发现一些问题与。nextline（）和。nextint（）有类似的问题，但不完全是这样。不过，如果真的是复制品，我提前道歉...

（如果需要，我会添加更多信息）

Answer 1

您正在调用nextint()，因此输入000被解析为数字0，因此int返回值为0。因此integer.toString()将返回字符串“0”。你为什么对此感到惊讶？

因为您实际上并不关心作为数字的输入，所以只需使用next:

for (int j = 0; j < row; j++) {
    String s = in.next();
    for (int k = 0; k < column; k++) {
        world[j][k] = Character.getNumericValue(s.charAt(k));
    }
}

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