Enter an integer n and find the number of occurrences of 1 in the decimal representation of n integers from 1 to n.
For example, if you enter 12, the numbers containing 1 in the integers from 1 to 12 are 1, 10, 11, and 12, and 1 appears 5 times in total.
Example 1:
Input: n = 12
Output: 5
Example 2:
Input: n = 13
Output: 6
limit:
1 <= n < 2^31
Because it is the number of occurrences of 1, it can be disassembled into 0~n, and how many 1s appear in each digit,
so continue dismantling. For example, 109.
For the ones digit: the
previous cycle is similar to 0~407, and 0~9 appears every time. 1 this 1, so a total of 407*1+ 0 ~ 9, which is 408
For the tens place:
40 times in the previous cycle, each time is 0~100, then for the tens place, there are 10 times,
40*10 + 0 = 400
For the hundreds place: 1 is a bit special, because it is 1 so the low digits should also be counted. For example, 14, then it is 0 10 + 4+1
5 100 + 09 +1=510
So the rule is drawn:
suppose the current position is a, the low position is b, the high position is c
a=0 当前位贡献： c*位数（个位就是1，十位10，百位100）
a=1 当前位贡献：c*位数+b+1
a>1 当前位贡献：c*位数+位数
class Solution {
public:
int countDigitOne(int n) {
long sum=0;
int b=0;
for(long i=1;n!=0;i*=10)
{
int a=n%10;
int c=n/10;
if(a==0)
{
sum+=c*i;
}
else if(a==1)
{
sum+=c*i+b+1;
}
else
{
sum+=c*i+i;
}
b=b+a*i;
n/=10;
}
return sum;
}
};