233 Number of Digit One

233 Number of Digit One

这道题是递归算法

class Solution:
    # @param {integer} n
    # @return {integer}
    def countDigitOne(self, n):
        if n <= 0:
            return 0
        if n < 10:
            return 1
        s = str(n)
        ls = len(s)
        ans = 0
        if s[0] == "1":
            ans += int(s[1:]) + 1 + pow(10, ls-2) * (ls-1)
            ans += self.countDigitOne(int(s[1:]))
        else:
            ans += int(s[0]) * pow(10, ls-2) * (ls-1) + pow(10, ls-1)
            ans += self.countDigitOne(int(s[1:]))
        return ans