[剑指 offer] JT31---the number of times 1 appears in an integer from 1 to n (everything is regular, especially numbers)

The topic is as follows

Idea and code

The first two are reckless solutions (the interview must be cool)

1. String solution

class Solution {
    
    
public:
    int NumberOf1Between1AndN_Solution(int n) {
    
    
        int ans=0;
        for(int j=1;j<=n;j++){
    
    
            string temp=to_string(j);
            for(int i=0;i<temp.length();i++){
    
    
                if(temp[i]=='1') ans++;
            }        
        }
        return ans;
    }
};

2. Numerical solution (similar to 1, except that the string is not converted)

public int NumberOf1Between1AndN_Solution(int n) {
    
    
    int count=0;
    for(int i=n;i>0;i--){
    
    
        for(int j=i;j>0;j/=10){
    
    
            if(j%10==1) count++;
        }
    }return count;

The j/=10 here is still very beautiful

3. Mathematical method (dream back to the college entrance examination)

Look at the code comments, you will understand

class Solution {
    
    
public:
    int NumberOf1Between1AndN_Solution(int n) {
    
    
        int count=0;//1的个数
        int i=1;//当前位
        int current=0,after=0,before=0;
        while((n/i)!=0){
    
    
            before=n/(i*10);//高位
            current=(n/i)%10;//当前位
            after=n-(n/i)*i;//低位
            if(current==0){
    
    
                //如果为0,由高位决定，等于高位数字*当前位数
                count+=before*i;
            }else if(current==1){
    
    
                //如果为1，出现1的次数由高位和低位决定，高位*当前位+低位+1
                count+=before*i+after+1;
            }else if(current>1){
    
    
                //如果大于1，出现1的次数由高位决定，(高位数字+1)*当前位数
                count+=(before+1)*i;
            }
            //前移一位
            i*=10;
        }
        return count;
    }
};