从1到n整数中1出现的次数.py

'''
题目描述
求出1~13的整数中1出现的次数,并算出100~1300的整数中1出现的次数？为此他特别数了一下1~13中包含1的数字有1、10、11、12、13因此共出现6次,但是对于后面问题他就没辙了。ACMer希望你们帮帮他,并把问题更加普遍化,可以很快的求出任意非负整数区间中1出现的次数（从1 到 n 中1出现的次数）。
参考：https://blog.csdn.net/yi_afly/article/details/52012593
'''

# -*- coding:utf-8 -*-
class Solution:
    def NumberOf1Between1AndN_Solution(self, n):
        # write code here
        if n<1:
            return 0
        count = 0
        base = 1
        round = n
        while round>0:
            weight = round%10;
            round/=10;
            count += round*base;
            if weight==1:
                count+=(n%base)+1;
            elif weight>1:
                count+=base;
            base*=10
        return count

#剑指offer解法
# -*- coding:utf-8 -*-
class Solution:
    def NumberOf1Between1AndN_Solution(self, n):
        # write code here
        if n <= 0:
            return 0

        return self.NumberOf1(str(n))
    def NumberOf1(self,strN):
        if not strN:
            return 0

        first = int(strN[0])
        length = len(strN)
        if length == 1 and first == 0:
            return 0
        if length == 1 and first>0:
            return 1
        numFristDigit=0
        if first>1:
            numFristDigit=10**(length-1)
        elif first == 1:
            numFristDigit = int(strN[1:])+1

        numOtherDigits = first*(length-1)*10**(length-2)
        numRecursive = self.NumberOf1(strN[1:])

        return numFristDigit+numOtherDigits+numRecursive