*LeetCode-Number of Digit One

每次累加一个位置上出现的1 循环每次除数成以10

每个位置上的1要看这一位这一位前面的数字 就是比如这一位是百 就要看有多少个百 

当百位是0和1的时候要特殊对待 其中1又要特殊对待

public class Solution {
public int countDigitOne(int n) {
int ones = 0;
for ( long m = 1; m <= n; m *= 10 ) {
int a = (int)(n / m);
int b = (int)(n % m);
if ( a % 10 == 1 )
ones += a / 10 * m + b + 1;
else if ( a % 10 == 0 )
ones += a / 10 * m;
else
ones += ( a / 10 + 1 ) * m;
}
return ones;
}
}
Go through the digit positions by using position multiplier
m with values 1, 10, 100, 1000, etc.

For each position, split the decimal representation into two parts, for example split n=3141592 into a=31415 and b=92 when we're at m=100 for analyzing the hundreds-digit. And then we know that the hundreds-digit of n is 1 for
prefixes "" to "3141", i.e., 3142 times. Each of those times is a streak, though. Because it's the hundreds-digit, each streak is 100 long. So
(a / 10 + 1) * 100 times, the hundreds-digit is 1.
Consider the thousands-digit, i.e., when m=1000. Then a=3141 and b=592. The thousands-digit is 1 for prefixes "" to "314", so 315 times. And each time is a streak of 1000 numbers. However, since the thousands-digit is a 1, the
very last streak isn't 1000 numbers but only 593 numbers, for the suffixes "000" to "592". So
(a / 10 * 1000) + (b + 1) times, the thousands-digit is 1.
The case distincton between the current digit/position being 0, 1 and >=2 can easily be done in one expression. With
(a + 8) / 10 you get the number of full streaks, and
a % 10 == 1 tells you whether to add a partial streak.

int countDigitOne(int n) {
int ones = 0;
for (long long m = 1; m <= n; m *= 10) {
int a = n/m, b = n%m;
ones += (a + 8) / 10 * m + (a % 10 == 1) * (b + 1);
}
return ones;
}