# Digit Counts ### Source - leetcode: [Number of Digit One | LeetCode OJ](https://leetcode.com/problems/number-of-digit-one/) - lintcode: [(3) Digit Counts](http://www.lintcode.com/en/problem/digit-counts/) ~~~ 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 leetcode 上的有点简单，这里以 Lintcode 上的为例进行说明。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)O(n \times L)O(n×L), L 为n 的最大长度，拆成字符数组，空间复杂度 O(L)O(L)O(L).