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.
class Solution {
public:
int countDigitOne(int n) {
int ones = 0;
for (long long m = 1; m <= n; m *= 10) {
ones += (n / m + 8) / 10 * m + (n / m % 10 == 1) * (n % m + 1);
}
return ones;
}
};
这道题我们先给出代码，在对其进行分析。代码的主体逻辑是从最低位开始向左依次分析。
因为对于每一位，有几种情况需要分开去进行讨论，即：
【LeetCode】233. Number of Digit One
原文：http://www.cnblogs.com/jdneo/p/5272364.html