Number of Digit One——LeetCode⑩

//原题链接https://leetcode.com/problems/number-of-digit-one/

• 题目描述

  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.
  

• 思路分析
  1.暴力法:对10求余，判断个位是否为1，然后除10依次判断//会超时
  2.优化：总结n每位数的规律，
  以n=123为例
     （1的个数：count
     当前位数上的权重：weight
     当前位数上的数字:now
     上一位数：last，例如按照从右到左，2的上一位数为3
     循环次数：round）
  个位数：3属于大于等于1，第十三个循环开始，则count = round + 1 = 12+1；
                若为120，即now为小于1，开始第十三个循环，则count = round=12；
  十位数及以上位数：以十位数为例，2大于1，则count = round*weight + weight=1*10+10
                                                          若为11X,即now为1，则count = round*weight + 1+last=1*10+1+x
                                                          若为10X,即now等于0，则count=round*weight=1*10
• 源码附录
   

  class Solution {
      public int countDigitOne(int n) {
          if(n<1){
              return 0;
          }
  
          int count = 1;
          for(int i=0;i<=n;i++){
              while(i>0){
                  if(i%10 == 1){
                      count ++;
                  }
                  i = i / 10;
            }
          }
           return count;
      }
  }

  class Solution {
      public int countDigitOne(int n) {
          if(n<1){
              return 0;
          }
  
          int count = 0;
          int round = n;
          int weight = 1;
          int now = 0 ;
          while(round>0){
              now = round%10;
              round = round/10;
              if(now==1){
                  count = count + (n%weight)+1;
              }
              else if(now>1){
                  count = count + weight;
              }
              count = count + round*weight;
              weight = weight*10;
          }
           return count;
      }
  }