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.
Solution:
For example '8192':
1-999 -> countDigitOne(999)
1000-1999 -> 1000 of 1s + countDigitOne(999)
2000-2999 -> countDigitOne(999)
.
.
7000-7999 -> countDigitOne(999)
8000-8192 -> countDigitOne(192)
Count of 1s : countDigitOne(999)*8 + 1000 + countDigitOne(192)
Noticed that, if the target is '1192':
Count of 1s : countDigitOne(999)*1 + (1192 - 1000 + 1) + countDigitOne(192)
(1192 - 1000 + 1) is the 1s in thousands from 1000 to 1192.
public int countDigitOne(int n) { if(n <= 0) return 0; if(n < 10) return 1; int base = (int)Math.pow(10, (n+"").length() - 1); int k = n / base; return countDigitOne(base - 1) * k + (k == 1 ? (n-base+1) : base) + countDigitOne(n % base); }
非递归的解法:
int countDigitOne(int n) { int result = 0; int digit = 1, num = n; while (num) { int mod = num % 10; int sign = mod > 0 ? 1 : 0; num /= 10; int a = num * digit; int b = sign * (mod == 1 ? n % digit + 1: digit); result += a + b; digit *= 10; } return result; }