Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.
For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.
Analysis:
fullCount(i): how many 1 there are for 0...10^i-1, e.g. fullCount(2) is the number of 1 in [0...99]
Now, we need count the 1s in n digit by digit. Using an eample, n=456, we count the 1s in [0...400] (for digit 4) + [0...50] (for digit 5) + [0...6] (for digit 6).
For digit 4, which belongs to 10^i where i=3, there are 4*fullCount(2), i.e., 1s in [0...99] for 000, 100, 200, 300, + the 100 1s on the current digit for [100...199].
We can use similar way to calculate digit 5 and digit 6.
If digit x > 1, then
digitCount(i,x) = x*fullCount(i-1) + 10^(i-1);
Specially, if digit x==1 then
digitCount(i,x) = 1* fullCount(i-1) + n%10^(i-1) + 1.
e.g., n=156, for digit 1, the count is fullCount(2) for [0...99] + (56+1) for [100...156].
On the other hand, fullCount(i) = 10*fullCount(i-1) + 10^(i-1),
which directly equals to i*10^(i-1).
Solution:
1 public class Solution { 2 public int countDigitOne(int n) { 3 int iter = 0 ; 4 int fullCount = 0; 5 long power10 = 1; 6 int res = 0; 7 while (n >= power10) { 8 int digit = n / (int) power10 % 10; 9 int digitCount = 0; 10 if (digit > 1) { 11 digitCount = digit * fullCount + (int) power10; 12 } else if (digit == 1) { 13 digitCount = fullCount + n % (int) power10 + 1; 14 } 15 res += digitCount; 16 17 // Get iter, power10, full count. 18 // IMPORTANT: the sequence here is important! 19 // because fullCount(i) = i * 10^(i-1). 20 iter++; 21 fullCount = iter * (int) power10; 22 power10 *= 10; 23 } 24 return res; 25 } 26 }