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Java-based LeetCode algorithm problem solutions, regularly updated
package g0201_0300.s0233_number_of_digit_one;
// #Hard #Dynamic_Programming #Math #Recursion
// #2022_07_04_Time_0_ms_(100.00%)_Space_41.2_MB_(25.50%)
@SuppressWarnings("java:S127")
public class Solution {
public int countDigitOne(int n) {
int ans = 0;
// count total number of 1s appearing in every digit, starting from the last digit
for (int k = n, cum = 0, curr10 = 1; k > 0; curr10 *= 10) {
int rem = k % 10;
int q = k / 10;
if (rem == 0) {
ans += q * curr10;
} else if (rem == 1) {
ans += q * curr10 + cum + 1;
} else {
ans += (q + 1) * curr10;
}
k = q;
// if loop is at 3rd last digit and n = 54321, cum is now = 321
cum += rem * curr10;
}
return ans;
}
}