Digit Counts

Source

Count the number of k's between 0 and n. k can be 0 - 9.

Example
if n=12, k=1 in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],
we have FIVE 1's (1, 10, 11, 12)

题解

找出从0至整数 n 中出现数位k的个数,与整数有关的题大家可能比较容易想到求模求余等方法,但其实很多与整数有关的题使用字符串的解法更为便利。将整数 i 分解为字符串,然后遍历之,自增 k 出现的次数即可。

Java

class Solution {
    /*
     * param k : As description.
     * param n : As description.
     * return: An integer denote the count of digit k in 1..n
     */
    public int digitCounts(int k, int n) {
        int count = 0;
        char kChar = (char)(k + '0');
        for (int i = k; i <= n; i++) {
            char[] iChars = Integer.toString(i).toCharArray();
            for (char iChar : iChars) {
                if (kChar == iChar) count++;
            }
        }

        return count;
    }
}

源码分析

太简单了,略

复杂度分析

时间复杂度 O(n×L), L 为n 的最大长度,拆成字符数组,空间复杂度 O(L).

 

posted @ 2021-08-22 14:16  凌雨尘  阅读(35)  评论(0编辑  收藏  举报