Digit Counts
Count the number of k's between 0 and n. k can be 0 - 9.
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)
分析:
利用while 循环 + num % 10可以获取 num中的所有数字。
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 totalCount = ;
for (int i = ; i <= n; i++) {
totalCount += counts(k, i);
}
return totalCount;
}
public int counts(int k, int value) {
int count = ;
if (value == && k == ) return ;
while (value != ) {
int remainder = value % ;
if (remainder == k) {
count++;
}
value = value / ;
}
return count;
}
};
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