Question

我有一个String数组

String[] list = new String[]{"2","A","B","1","C","3","D","E","F"}

请注意，数组的第一个元素是2。我会做的 -

int n = Integer.parseInt(list[0]);

问题是，我需要一个循环来完成数组。我似乎无法解析它而没有给我NumberFormatException错误。另外，请注意，如果数字是2，它将排列接下来的2个元素（可能是随机的），然后是空格。

1：A

2：B

3：-empty -

4：D

5：E

6：F

7：-empty -

8：C

Answer 1

遍历数组并检查所选字符是否为数字。如果是数字使用parseInt。

这是一个如何识别数字的教程。 What is the best way to tell if a character is a letter or number in Java without using regexes?

祝你好运。

Answer 2

迭代数组时，在循环中使用try / catch语句。

这是一个简短的例子：

for(int i = 0; i < list.length; i++) {
    try {
        int n = Integer.parseInt(list[i]);
        // do whatever you need to do with n here
    }
    catch (NumberFormatException nfex) {
        // do something else with the current element if need be
    }
}

Answer 3

for(String hexNbr : list) 
{
   System.out.println("Next number is " + Integer.parseInt(hexNbr, 16);
}

您的输入是十六进制，因此您必须使用parseInt（String，radix）。

Answer 4

算法很简单。

解析数
将N个值左移1个位置
将空值分配给空置空间
提前过去，然后重复

进展如下：

Start: "2","A","B","1","C","3","D","E","F"
N = 2
Shift: "A","B","B","1","C","3","D","E","F"
Empty: "A","B","-","1","C","3","D","E","F"
N = 1
Shift: "A","B","-","C","C","3","D","E","F"
Empty: "A","B","-","C","-","3","D","E","F"
N = 3
Shift: "A","B","-","C","-","D","E","F","F"
Empty: "A","B","-","C","-","D","E","F","-"

代码的

String[] list = {"2","A","B","1","C","3","D","E","F"};
for (int i = 0; i < list.length; ) {
    int len = Integer.parseInt(list[i]);
    System.arraycopy(list, i + 1, list, i, len);
    list[i + len] = "-empty-";
    i += len + 1;
}
for (String value : list)
    System.out.println(value);

输出

A
B
-empty-
C
-empty-
D
E
F
-empty-

是否可以将Integer.parse用于“混合”类型的String数组？

4 个答案: