最高赞的解法果然牛逼

class Solution {
    public int numberOf1Between1AndN_Solution(int n) {
        return IntStream
                .iterate(1, x -> x * 10)
                .limit(String.valueOf(n).length())
                .map(x -> (n / x + 8) / 10 * x + ((n / x) % 10 == 1 ? n % x + 1 : 0))
                .sum();
    }
}

暴力写法 最后一个数据直接过不了：

class Solution {
    public int numberOf1Between1AndN_Solution(int n) {
        return IntStream
                .rangeClosed(1, n)
                .mapToObj(String::valueOf)
                .mapToInt(x -> (int) IntStream
                        .range(0, x.length())
                        .map(x::charAt)
                        .filter(val -> val == '1')
                        .count())
                .sum();
    }
}