先把数字转成string 写个自定义排序就行
我写了两种比较两个string的大小
直接相加再比较数字 较慢

@Override
public int compare(String a, String b) {
    int o1 = Integer.parseInt(a+b);
    int o2 = Integer.parseInt(b+a);
    return o1 - o2;
}

手动一位一位比较 较快

@Override
public int compare(String a, String b) {
    int ax = 0, bx = 0, av = 0, bv = 0;
    while (ax*bx < 2 && a.charAt(av) == b.charAt(bv)){
        if (++av == a.length()) {av = 0;++ax;}
        if (++bv == b.length()) {bv = 0;++bx;}
    }
    return a.charAt(av) - b.charAt(bv);
}

stream写法 直接return 但是stream对string处理速度慢一些

class Solution {
    public String printMinNumber(int[] nums) {
       return nums.length == 0 ? "" :
                Arrays.stream(nums)
                .mapToObj(String::valueOf)
                .sorted((a,b) -> {
                    int ax = 0, bx = 0, av = 0, bv = 0;
                    while (ax*bx < 2 && a.charAt(av) == b.charAt(bv)){
                        if (++av == a.length()) {av = 0;++ax;}
                        if (++bv == b.length()) {bv = 0;++bx;}
                    }
                    return a.charAt(av) - b.charAt(bv);
                })
                .reduce((a,b) -> a+b)
                .get();
    }
}

正常写法 速度几乎提升一倍

class Solution {
    public String printMinNumber(int[] nums) {
       String[] strings = new String[nums.length];
        for (int i = 0; i < nums.length; ++i) {
            strings[i] = String.valueOf(nums[i]);
        }
        Arrays.sort(strings, new Comparator<String>() {
            @Override
            public int compare(String a, String b) {
                int av = 0, bv = 0, ax = 0, bx = 0;
                while (ax*bx < 2 && a.charAt(av) == b.charAt(bv)){
                    if (++av == a.length()) {av = 0;++ax;}
                    if (++bv == b.length()) {bv = 0;++bx;}
                }
                return a.charAt(av) - b.charAt(bv);
            }
        });
        StringBuilder res = new StringBuilder();
        for (String str : strings) {
            res.append(str);
        }
        return res.toString();
    }
}