标签：题解 i++ strong bsp ntc public nbsp lin ==
题目：
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)
题解：
Solution 1 ()
class Solution { public: int digitCounts(int k, int n) { if (n < 0) { return 0; } int cnt = 0; for (int i = 1; i <= n; i++) { int num = i; while (num) { if (num % 10 == k) { cnt++; } num = num / 10; } } if (k == 0 && n >= 0) { cnt++; } return cnt; } };
标签：题解 i++ strong bsp ntc public nbsp lin ==
原文地址：http://www.cnblogs.com/Atanisi/p/6880570.html
