LeetCode - Number of Digit One
Problem description
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.
Example:
1
2
3
Input: 13
Output: 6
Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.
Analysis
This is a standard digit dp.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
class Solution {
public int countDigitOne(int n) {
if (n <= 0){
return 0;
}
int temp = n;
int digits = 0;
while(temp != 0){
digits++;
temp = temp / 10;
}
if (digits == 1){
return 1;
}
int[] dp = new int[digits+1];
dp[0] = 0;
for (int i = 1; i < digits+1; i++){
dp[i] = dp[i-1] * 10 + (int)Math.pow(10, i-1);
}
int pow = (int)Math.pow(10, digits-1);
int p = digits;
int res = 0;
while (n > 0 && p >= 1){
if (n >= pow){
int dividend = n/pow;
n = n % pow;
if (dividend == 1){
res += dp[p-1] + n + 1;
}
else{
res += dp[p-1] * dividend + pow;
}
}
pow /=10;
p--;
}
return res;
}
}
What to improve
- be careful about the +1 part in dp array.