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Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.
For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.
Hint:
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case 1: n=3141092, a= 31410, b=92. 计算百位上1的个数应该为 3141 *100 次.
case 2: n=3141192, a= 31411, b=92. 计算百位上1的个数应该为 3141 *100 + (92+1) 次.
case 3: n=3141592, a= 31415, b=92. 计算百位上1的个数应该为 (3141+1) *100 次.
以上三种情况可以用 一个公式概括:
(a + 8) / 10 * m + (a % 10 == 1) * (b + 1);class Solution { public: int countDigitOne(int n) { int ones = 0; for (long long i = 1; i <= n; i *= 10) { ones += (n / i + 8) / 10 * i + (n / i % 10 == 1) * (n % i + 1); } return ones; } };
[LeetCode]67. Number of Digit One1的个数和
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原文地址：http://www.cnblogs.com/aprilcheny/p/4949316.html
