233. Number of Digit One

public class Solution {    public int countDigitOne(int n) {        if(n<=0)return 0;        return dfs(n);    }    private int dfs(int n){        int length = size(n);        if(length==1){            if(n==0)return 0;            else{                return 1;            }        }        if(length>1){            int count=0;//首位一的个数            int len = (n-(shengcheng(length-1)+1))/(shengcheng(length-1)+1);//shengcheng(length-1)+1,其实就是生成整跟n相同位数的整十数，如10，100，1000，10000000，            int firstNum = n/(shengcheng(length-1)+1);            int behindNum = n-firstNum*(shengcheng(length-1)+1);            if (len>0)count=shengcheng(length-1)+1;            else {                count = n-(shengcheng(length-1)+1)+1;            }            return dfs(shengcheng(length-1))+len*dfs(shengcheng(length-1))+count+dfs(behindNum);        }//比如321=dfs(99)+2*dfs(99)+100+dfs(21),dfs(1092)=dfs(999)+0*dfs(999)+92+dfs(92),那321来看最合适        return 0;    }    public int size(int n){//测量数的位数        int count = 1;        while (n/10>0){            n = n/10;            count++;        }        return count;    }    public int shengcheng(int n){//生成99.999.9999这样的数三位就是999,4位就是9999        int count=1;        for(int i=0;i<n;i++)            count  = count*10;        return count-1;    }}