题目

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:

思路

代码

/*---------------------------------------*   日期：2015-07-19*   作者：SJF0115*   题目: 233.Number of Digit One*   网址：https://leetcode.com/problems/number-of-digit-one/*   结果：AC*   来源：LeetCode*   博客：-----------------------------------------*/#include 
     
      #include 
      
       using namespace std;class Solution {public:    int countDigitOne(int n) {        if(n == 0){            return 0;        }//if        int result = 0;        int lowerNum = 0,curNum = 0,highNum = 0;        int base = 1;        int num = n;        while(num){            // 低位部分            lowerNum = n - num * base;            // 当前部分            curNum = num % 10;            // 高位部分            highNum = num / 10;            // 如果为0则这一位1出现的次数由更高位决定 (更高位数字*当前位数)            if(curNum == 0){                result += highNum * base;            }//if            // 如果为1则这一位1出现的次数不仅受更高位影响还受低位影响(更高位数字*当前位数+低位数字+1)            else if(curNum == 1){                result += highNum * base + (lowerNum + 1);            }//else            // 大于1则仅受更高位影响((更高位数字+1)*当前位数)            else{                result += (highNum + 1) * base;            }//else            num /= 10;            base *= 10;        }//while        return result;    }};int main(){    Solution s;    int n;    while(cin>>n){        cout<
       
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