【leetcode】233. Number of Digit One

最新推荐文章于 2025-09-07 20:32:33 发布

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最新推荐文章于 2025-09-07 20:32:33 发布

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原文链接：http://www.cnblogs.com/seyjs/p/10943714.html

博客围绕给定整数n，计算小于等于n的所有非负整数中数字1出现的总个数这一问题展开。给出解题思路，称其为《编程之美》上的题目，还给出了相关代码，代码转载自特定博客。

题目如下：

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

Example:
Input: 13
Output: 6 
Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

解题思路：本题是《编程之美》上的题目，详细分析如下图。

代码如下：

class Solution(object):
    def countDigitOne(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n <= 0:
            return 0
        res = 0
        sn = str(n)
        length = len(sn)
        inx = len(sn) - 1
        while inx >= 0:
            lower = sn[inx+1:]
            current = sn[inx]
            higher = sn[:inx]
            if int(current) == 0:
                if len(higher) > 0:
                    res += int(higher) * (10 ** (length - inx - 1))
            elif int(current) == 1:
                if len(higher) > 0:
                    res += int(higher) * (10 ** (length - inx - 1))
                if len(lower) > 0:
                    res += int(lower)
                res += 1

            else:
                if len(higher) > 0:
                    res += (int(higher))  * (10 ** (length - inx - 1))
                res += 1 * (10 ** (length - inx - 1))
            inx -= 1
        return res