Question

如果我有String x =＆＃34; 2016＆＃34;如何将其转换为int类型的6210？

我正在尝试

String x = "2016"
int v = 0;
for (int k = 0; k < x.length(); k++) {
    if (v < x.charAt(k) - '0') {
        v = x.charAt(k) - '0';
    }
}
v *= 1000;

但我不知道下一步该做什么

Answer 1

char[] stringVals = x.toCharArray();
int[] results = new int[stringVals.length];
for (int i = 0; i < stringVals.length; i++) {
        try {
            results[i] = Integer.parseInt(stringVals[i]);
        } catch (NumberFormatException nfe) {
            //NOTE: write something here if you need to recover from formatting errors
        };
    }
 Arrays.sort(results);
 String sortedResult = Arrays.toString(results);

Answer 2

尝试查看Integer.parseInt

但你不能使用chars（作为他们答案中提到的前一张海报）因为char与Integer.parseInt不兼容，只有字符串。

尝试将String解析为如下数组：

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

潜在的解决方案：

String x = "2016";

    String[] numbers = x.split("");
    Integer[] nums = new Integer[numbers.length];
    for (int k = 0; k < numbers.length; k++) {

        nums[k] = Integer.parseInt(numbers[k]);
                }

    Arrays.sort(nums, Collections.reverseOrder());
    StringBuilder strNum = new StringBuilder();

    for (int num : nums){
        strNum.append(num);
        }
    int finalInt = Integer.parseInt(strNum.toString());
    System.out.println(finalInt);

Answer 3

使用java-8

String reversed = Arrays.stream(x.split(""))
                        .sorted(String.CASE_INSENSITIVE_ORDER.reversed())
                        .collect(Collectors.joining());

int v = Integer.parseInt(reversed);

如何将一串数字转换为排序的int？

3 个答案: