Question

Formatted question description: https://leetcode.ca/all/233.html

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:

Input: n = 13
Output: 6


Example 2:

Input: n = 0
Output: 0


Constraints:

• 0 <= n <= 109

Algorithm

This question is actually equivalent to a question to find a pattern. So in order to find the rule, we first enumerate all the numbers containing 1 and count the number every 10

 Number of 1 Numbers with 1 Number range 1 1 [1, 9] 11 10 11 12 13 14 15 16 17 18 19 [10, 19] 1 21 [20, 29] 1 31 [30, 39] 1 41 [40, 49] 1 51 [50, 59] 1 61 [60, 69] 1 71 [70, 79] 1 81 [80, 89] 1 91 [90, 99] 11 100 101 102 103 104 105 106 107 108 109 [100, 109] 21 110 111 112 113 114 115 116 117 118 119 [110, 119] 11 120 121 122 123 124 125 126 127 128 129 [120, 129]

There are only 1 1 within 100 except for 11 ‘1’s between 10-19.

If you don’t consider the 10 extra ‘1’s in the interval [10, 19], then for any two-digit number, the number on the ten’s digit (plus 1) represents the number of occurrences of 1.

Code

• 
public class Number_of_Digit_One {

class Solution {
public int countDigitOne(int n) {
int res = 0;
for (long k = 1; k <= n; k *= 10) {
long r = n / k, m = n % k;
res += (r + 8) / 10 * k + (r % 10 == 1 ? m + 1 : 0);
}
return res;
}
}
}

############

class Solution {
private int[] a = new int[12];
private int[][] dp = new int[12][12];

public int countDigitOne(int n) {
int len = 0;
while (n > 0) {
a[++len] = n % 10;
n /= 10;
}
for (var e : dp) {
Arrays.fill(e, -1);
}
return dfs(len, 0, true);
}

private int dfs(int pos, int cnt, boolean limit) {
if (pos <= 0) {
return cnt;
}
if (!limit && dp[pos][cnt] != -1) {
return dp[pos][cnt];
}
int up = limit ? a[pos] : 9;
int ans = 0;
for (int i = 0; i <= up; ++i) {
ans += dfs(pos - 1, cnt + (i == 1 ? 1 : 0), limit && i == up);
}
if (!limit) {
dp[pos][cnt] = ans;
}
return ans;
}
}

• class Solution:
def countDigitOne(self, n: int) -> int:
@cache
def dfs(pos, cnt, limit):
if pos <= 0:
return cnt
up = a[pos] if limit else 9
ans = 0
for i in range(up + 1):
ans += dfs(pos - 1, cnt + (i == 1), limit and i == up)
return ans

a = [0] * 12
l = 1
while n:
a[l] = n % 10
n //= 10
l += 1
return dfs(l, 0, True)

############

class Solution(object):
def countDigitOne(self, n):
"""
:type n: int
:rtype: int
"""
m = 1
ones = 0
while m <= n:
r = (n / m) % 10
if r > 1:
ones += m
elif r == 1:
ones += n % m + 1

ones += (n / (m * 10)) * m
m *= 10
return ones


• class Solution {
public:
int a[12];
int dp[12][12];

int countDigitOne(int n) {
int len = 0;
while (n) {
a[++len] = n % 10;
n /= 10;
}
memset(dp, -1, sizeof dp);
return dfs(len, 0, true);
}

int dfs(int pos, int cnt, bool limit) {
if (pos <= 0) {
return cnt;
}
if (!limit && dp[pos][cnt] != -1) {
return dp[pos][cnt];
}
int ans = 0;
int up = limit ? a[pos] : 9;
for (int i = 0; i <= up; ++i) {
ans += dfs(pos - 1, cnt + (i == 1), limit && i == up);
}
if (!limit) {
dp[pos][cnt] = ans;
}
return ans;
}
};

• func countDigitOne(n int) int {
a := make([]int, 12)
dp := make([][]int, 12)
for i := range dp {
dp[i] = make([]int, 12)
for j := range dp[i] {
dp[i][j] = -1
}
}
l := 0
for n > 0 {
l++
a[l] = n % 10
n /= 10
}
var dfs func(int, int, bool) int
dfs = func(pos, cnt int, limit bool) int {
if pos <= 0 {
return cnt
}
if !limit && dp[pos][cnt] != -1 {
return dp[pos][cnt]
}
up := 9
if limit {
up = a[pos]
}
ans := 0
for i := 0; i <= up; i++ {
t := cnt
if i == 1 {
t++
}
ans += dfs(pos-1, t, limit && i == up)
}
if !limit {
dp[pos][cnt] = ans
}
return ans
}
return dfs(l, 0, true)
}

• public class Solution {
public int CountDigitOne(int n) {
if (n <= 0) return 0;
if (n < 10) return 1;
return CountDigitOne(n / 10 - 1) * 10 + n / 10 + CountDigitOneOfN(n / 10) * (n % 10 + 1) + (n % 10 >= 1 ? 1 : 0);
}

private int CountDigitOneOfN(int n) {
var count = 0;
while (n > 0)
{
if (n % 10 == 1) ++count;
n /= 10;
}
return count;
}
}