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:
class Solution { public: int countDigitOne(int n) { if(n <= 0) return 0; vector<int> v; int t = n; while(t) { v.push_back(t%10); t /= 10; } int l = v.size(), rest = n - v[l-1] * pow(10, l-1); return (v[l-1] > 1 ? pow(10, l-1) : rest + 1) + v[l-1] * (l-1) * pow(10, l-2) + countDigitOne(rest); } };
233. Number of Digit One *HARD* -- 从1到n的整数中数字1出现的次数
原文：http://www.cnblogs.com/argenbarbie/p/5827992.html