题解 | #把数组排成最小的数#

class Solution {
    public String PrintMinNumber(int[] numbers) {
        String str = "";
        //冒泡排序
        for (int i = 0; i < numbers.length; i++) {
            //注意比较规则，先拼接起来，再比较
            for (int j = i + 1; j < numbers.length; j++) {
                int a = Integer.valueOf(numbers[i] + "" + numbers[j]);
                int b = Integer.valueOf(numbers[j] + "" + numbers[i]);
                if (a > b) {
                    int t = numbers[i];
                    numbers[i] = numbers[j];
                    numbers[j] = t;
                }

            }
        }
        //拼接排好序的数组
        for (int s:numbers) {
            str += s;
        }
        return str;
    }
}

和字符串拼接之后比较的思想类似，

这里也是比较两个数值大小时，需要比较拼接之后的数字

或者用这种

  StringBuffer str = new StringBuffer();
        ArrayList<Integer> list = new ArrayList<>();
        for (int i = 0; i < arr.length; i++) {
            list.add(arr[i]);
        }

        Collections.sort(list, new Comparator<Integer>() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return o1 - o2;
            }
        });
        System.out.println(list);


        return str.toString();