LeetCode 233 Number of Digit One 某一范围内的整数包含1的数量

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.

方法一

 class Solution {

 public:

     int countDigitOne(int n) {

         int cnt=;

         for(long long m=;m<=n;m*=)

         {

             int a=n/m,b=n%m;

             if(a%==)

                 cnt+=a/*m;

             else if(a%==)

                 cnt+=a/*m+(b+);

             else

                 cnt+=(a/+)*m;

         }

         return cnt;

     }

 };

方法二

 class Solution {

 public:

     int countDigitOne(int n) {

         int cnt=;

         for(long long m=;m<=n;m*=)

             cnt=cnt+(n/m+)/*m+(n/m%==)*(n%m+);

         return cnt;

     }

 };

巴特西

LeetCode 233 Number of Digit One 某一范围内的整数包含1的数量

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