我需要帮助重命名数组中的元素

Question

我在学校做一个 java 项目，我们遇到了以下问题：

rename – 更改现有工作表的名称

public int rename(String currentName, String newName)

rename 传递了两个工作表名称。仅当 currentName 工作表名称在列表中而 newName 工作表名称不在时，才应执行重命名。否则 rename 什么都不做（即列表不受影响）。 如果currentName 成功更改为newName，则该方法返回重命名的工作表的索引位置。否则返回-1。

这是我目前所拥有的

public int rename(String currentName, String newName) {
    int i = 0;
    for (i=0; i<SheetsNames.length; i++) {
        if (SheetsNames[i].equals(currentName)) {
            SheetsNames[i] = newName;
                return i;
        }
    }
    return -1;
}

我得到了重命名部分，但我无法让它不重命名为具有相同名称的东西

Answer 1

这是解决方案。

1) 验证输入参数

2) 在数组中查找字符串并返回索引

public class SheetNames {

    public Integer changeSheetName(String[] sheetNames, String currentName, String newName) {

        if (sheetNames == null || sheetNames.length == 0 || currentName == null || newName == null) {
            throw new IllegalArgumentException("Invalid input");
        }

        Integer replacedIndex = Arrays.asList(sheetNames).indexOf(currentName);
        if (replacedIndex != -1) {
            sheetNames[replacedIndex] = newName;
        }
        System.out.println(Arrays.toString(sheetNames));
        return replacedIndex;
    }
}

Junit 测试：-

public class SheetNamesTest {

    private SheetNames sheetNames;

    @Before
    public void setup() {
        sheetNames = new SheetNames();
    }

    @Test
    public void shouldReplaceSheetName() {
        String[] sheetNamesArr = {"dennis", "john", "dan"};
        Assert.assertEquals(new Integer(1), sheetNames.changeSheetName(sheetNamesArr, "john", "peter"));

        String[] sheetNamesArr2 = {"john", "dan"};
        Assert.assertEquals(new Integer(0), sheetNames.changeSheetName(sheetNamesArr2, "john", "peter"));
    }

    @Test
    public void shouldNotReplaceSheetName() {
        String[] sheetNamesArr = {"dennis", "johnhunt", "dan"};
        Assert.assertEquals(new Integer(-1), sheetNames.changeSheetName(sheetNamesArr, "john", "peter"));
    }

    @Test(expected = IllegalArgumentException.class)
    public void shouldThrowExceptionWhenSheetNameIsNull() {
        sheetNames.changeSheetName(null, "john", "peter");
    }

    @Test(expected = IllegalArgumentException.class)
    public void shouldThrowExceptionWhenSheetNameIsEmpty() {
        String[] sheetNamesArr = {};
        sheetNames.changeSheetName(sheetNamesArr, "john", "peter");
    }

    @Test(expected = IllegalArgumentException.class)
    public void shouldThrowExceptionWhenCurrentNameIsNull() {
        String[] sheetNamesArr = {"dennis", "johnhunt", "dan"};
        sheetNames.changeSheetName(sheetNamesArr, null, "peter");
    }

    @Test(expected = IllegalArgumentException.class)
    public void shouldThrowExceptionWhenNewNameIsNull() {
        String[] sheetNamesArr = {"dennis", "johnhunt", "dan"};
        sheetNames.changeSheetName(sheetNamesArr, "john", null);
    }

}

Answer 2

阅读下面代码中的cmets：

public int rename(String currentName, String newName) {
    int currentNameindex = -1; // set the initial position of currentName

    for (int i = 0; i < SheetsNames.length; i++) {
        if (currentNameindex < 0)  // currentName has not been found yet 
            if (SheetsNames[i].equals(currentName))
                currentNameindex = i;  // save the position of currentName 

        if (SheetsNames[i].equals(newName))  // newName is found in the list
            return -1;  
    }

    if (currentNameindex > 0)  // currentName is found in the list
        SheetsNames[currentNameindex] = newName;  // rename the list item

    return currentNameindex;
}

Answer 3

遍历 SheetNames 并将 currentName 索引的任何实例存储在名为 index 的变量中，同时检查 newName 是否存在于列表中。如果有 newName 的实例自动返回 -1。但是，如果列表中没有 newName，则 currentName 的索引实例将存储在索引变量中（如果存在）。如果没有，则该函数无论如何都会返回 -1，因为这是索引初始化的内容。

public int rename(String currentName, String newName) {
    int index = -1;
    for (int i=0; i<SheetsNames.length; i++) {
        if (SheetsNames[i].equals(newName))
            return -1;
        if (SheetsNames[i].equals(currentName))
            index = i;
    }
    if (index != -1)
        SheetNames[index] = newName;
    return index; 
}

Answer 4

遍历您的 SheetsNames 并检查它是否已经包含 newName（如果是，那么您不想更新，因此返回 -1）

for(String name: SheetsNames){
   if(name.equals(newName) return -1;
}

现在我们确定新名称不在列表中！让我们检查其中是否有与 currentName 相同的名称，如果是则返回其索引。

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

   if(name.equals(currentName){
     SheetsNames[i] = newName;
     return i;
   }
}
return -1; //its not in SheetsNames, so we return -1