[LeetCode] Number of Digit One

The following idea is taken from a book named 《剑指offer》 published in China.

Suppose

n = 271

, it then breaks

[1, 271]

into

[1, 71]

and

[72, 271]

. For

[72, 271]

, the number of

1

on the hundreds are

10^2 = 100

(note that if the digit on the higest bit is

1

, then the number of

1

's on that bit will be

72

, which is

271 % 100 + 1

). To compute the number of

1

on tens and units, we further break

[72, 271]

into

[72, 171]

and

[172, 271]

. Each interval has

10^1 = 10

1

's on both the tens and units. So the overall number of

1

's on tens and units is

2 * 2 * 10^1 = 40

. Now we are only left with

[1, 71]

, which can be solved recursively. The

int n

is transformed into a

string num

to facilitate some operations.

The code is as follows.

class Solution {
public:
int countDigitOne(int n) {
if (n <= 0) return 0;
string num = to_string(n);
return countOne(num);
}
private:
int countOne(string num) {
if (num.empty()) return 0;
int first = num[0] - '0';
if (num.length() == 1) {
if (first) return 1;
return 0;
}
int highestBit;
if (first > 1) highestBit = powerTen(num.length() - 1);
else if (first == 1) highestBit = stoi(num.substr(1)) + 1;
int otherBits = first * (num.length() - 1) * powerTen(num.length() - 2);
int recurCount = countOne(num.substr(1));
return highestBit + otherBits + recurCount;
}
int powerTen(int exp) {
int p = 1;
for (int i = 0; i < exp; i++)
p *= 10;
return p;
}
};

This link has a brilliant 4-line solution! The code is rewritten as follows.

class Solution {
public:
int countDigitOne(int n) {
int counts = 0;
for (int m = 1000000000; m; m /= 10)
counts += (n / m + 8) / 10 * m + (n / m % 10 == 1) * (n % m + 1);
return counts;
}
};