java - java将数组值左移

Question

这是我关于同一方法的第二篇文章。该方法说明如下:

“将 2 个或更多空格的所有序列减少到 1 个空格 字符数组。如果删除任何空格则相同的数字 空字符 '\u0000' 将填充末尾的元素 数组。”

该方法的参数是一个字符数组。我已经成功地计算了重复空格的数量，但是，我一生都无法弄清楚如何在考虑此类重复空格的同时向下移动值。最后，该方法应该用“\u0000”字符替换重复空格的索引数量。这是我到目前为止所拥有的:

// Calculate duplicate count before changing the array
    int duplicateCount = 0;
    for(int i = 0; i + 1 < characters.length; i++){
        if(characters[i] == ' ' && characters[i + 1] == ' '){
            duplicateCount++;
        }
    }

    // Shift the array down however much is needed
    for(int j = 0; j + 1 < characters.length; j++){
        if(characters[j] == ' ' && characters[j + 1] == ' '){
            for(int a = j, b = a + 1; b < characters.length; a++, b++){
                characters[a] = characters[b];
            }
        }
    }
    for(int replace = characters.length - duplicateCount; replace < characters.length; replace++){
        characters[replace] = '\u0000';
    }
}

因此，如果输入为:char [] 个字符 = {'a', 't', ' ', '.', ' ', ' ', 'g', ' ', ' ', 'h', ' '};

输出应为:char [] ExpectedResult = {'a', 't', ' ', '.', ' ', 'g', ' ', 'h', ' ','\u0000', '\u0000'};

谢谢大家，这个问题看起来应该很简单，但我却被困住了。如果您能为您的答案提供任何解释，我将非常感激。再次感谢您。

Answer 1

就地完成这件事很容易。只需迭代数组，跟踪您正在检查的索引(以查看它是否是额外的空间)和您要复制到的索引。最后，用您的“\u0000”值填充数组即可完成。

我将其设置为一个简单的状态机，以便轻松跟踪我们是否获得了额外的空间。

public void squeezeMe(char[] characters) {
    SqueezeState state = SqueezeState.START;

    int p = 0;
    for (int i = 0; i < characters.length; i++) {
        SqueezeState newState = SqueezeState.START;

        // Evaluate input based on current state
        switch (state) {
        case START:
        case NOT_A_SPACE: {
            if (Character.isWhitespace(characters[i])) {
                newState = SqueezeState.FIRST_SPACE;
            } else {
                newState = SqueezeState.NOT_A_SPACE;
            }
            break;
        }
        case FIRST_SPACE:
        case EXTRA_SPACE: {
            if (Character.isWhitespace(characters[i])) {
                newState = SqueezeState.EXTRA_SPACE;
            } else {
                newState = SqueezeState.NOT_A_SPACE;
            }
        }
        }

        // Transition to new state
        switch (newState) {
        case NOT_A_SPACE:
        case FIRST_SPACE: {
            if (i > p) {
            characters[p] = characters[i];
            }
            p++;
            break;
        }
        }

        state = newState;
    }

    for (int i = p; i < characters.length; i++) {
        characters[i] = '\u0000';
    }

}

private enum SqueezeState {
    START, NOT_A_SPACE, FIRST_SPACE, EXTRA_SPACE;
}

@Test
public void test1() {
    char[] result = { 'a', 't', ' ', '.', ' ', ' ', 'g', ' ', ' ', 'h', ' ' };
    char[] expected = { 'a', 't', ' ', '.', ' ', 'g', ' ', 'h', ' ', '\u0000', '\u0000' };
    squeezeMe(result);
    assertEquals(expected.length, result.length);
    for (int i = 0; i < expected.length; i++) {
        assertEquals("Index " + i, expected[i], result[i]);
    }
}

如果您不想使用状态机，可以这样做:

public void squeezeMe(char[] characters) {
    boolean copyThis = false;
    boolean wasLastASpace = false;

    int p = 0;
    for (int i = 0; i < characters.length; i++) {
        if (Character.isWhitespace(characters[i])) {
            copyThis = !wasLastASpace;
            wasLastASpace = true;
        } else {
            copyThis = true;
            wasLastASpace = false;
        }

        if (copyThis) {
            if (i != p) {
                characters[p] = characters[i];
            }
            p++;
        }
    }

    for (int i = p; i < characters.length; i++) {
        characters[i] = '\u0000';
    }
}