# Digit Counts

## Question

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。

### C++

class Solution {
public:
/*
* param k : As description.
* param n : As description.
* return: How many k's between 0 and n.
*/
int digitCounts(int k, int n) {
char c = k + '0';
int count = 0;
for (int i = k; i <= n; i++) {
for (auto s : to_string(i)) {
if (s == c) count++;
}
}
return count;
}
};


### 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;
}
}


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