🔴 剑指 Offer 43. 1～n 整数中 1 出现的次数

LeetCode 提示

题目难度 困难

原题链接 🔗 leetcode

题解 1.抄的.py

# 太妙了# https://leetcode-cn.com/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-lcof/solution/1n-zheng-shu-zhong-1-chu-xian-de-ci-shu-umaj8/class Solution:    def countDigitOne(self, n: int) -> int:        dight, res = 1, 0        high, cur, low = n // 10, n % 10, 0
        while high != 0 or cur != 0:            if cur == 0:                res += high * dight            elif cur == 1:                res += high * dight + low + 1            else:                res += (high + 1) * dight                        low += cur*dight            cur = high % 10            high = high // 10            dight *= 10                return res