[LeetCode]67. Number of Digit One1的个数和

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:

1. Beware of overflow.

Subscribe to see which companies asked this question

 

解法：参考http://blog.csdn.net/xudli/article/details/46798619。从个位循环到最高位，每次计算某一位上是1的数字个数。以算百位上1为例子:   假设百位上是0, 1, 和 >=2 三种情况: 

    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;
    }
};