The number of occurrences of 1 in an integer ranging from 1 to n of offer (32)

Question: enter an integer to calculate the number of occurrences of 1 in the decimal representation of the N integers from 1 to n.

Method 1: The most intuitive solution T (n) = O (nlgn)

int NumberOf1Between1AndN_Solution1(unsigned int n){    int number = 0;    for(unsigned int i = 1; i <= n; ++ i)        number += NumberOf1(i);    return number;}int NumberOf1(unsigned int n){    int number = 0;    while(n)    {        if(n % 10 == 1)            number ++;        n = n / 10;    }    return number;}

 

Method 2:

Assume n = ABCDE (in decimal notation of N, E is a single digit)

If you want to calculate the number of times that a hundred bits of C appears 1, it will be affected by three factors:

(1) Hundreds of digits

(2) digits below a hundred bits (low bits)

(3) A hundred or more digits (high digit)

 

If the number of hundred digits is 0, the number of times that the hundred digits appear 1 depends on its high digit.

For example, 12013, 1 may be: 100-199 1100-1199 2100-2199 ...... 10100-10199 11100-11199

We can see that when 1 is displayed, the low position (CDE) is 100-199, while the high position (I .e., the AB two) ranges from 0, 1, 2, 3... 11 is exactly 12 pairs, that is, 12 high numbers

So: the number of times that a hundred bits appear 1: The high number 12 × the current number of BITs 100 = 1200

 

If the number of digits is 1, the number of times the number of digits appears 1 depends on the number of digits in the upper and lower positions.

For example, 12113, first according to the above situation: 100-199 1100-1199 2100-2199 ...... 10100-10199 11100-11199

At this time, if the number of BITs is 1, there may be another situation: 12101 12102 12103 ..... 12113 (note that both the height and the height are 12, and the low position ranges from 100 to 113, that is, 13 + 1 = 14 cases)

So: The number of times when a hundred digits appear 1: 12 x the current number of digits: 100 + 13 + 1 = 1214

 

If a hundred-digit number is greater than 1, the number of times a hundred-digit number appears 1 depends on its high-digit number.

For example, 12213, as described in the first case: 100-199 1100-1199 2100-2199 ...... 10100-10199 11100-11199 12100-12199

Note: A hundred-digit number is greater than 1, so do not forget the 12100-12199 case.

So: the number of times that a hundred bits appear 1: (12 + 1) x the current number of BITs 100 = 1300

 

Int onenum (const int num) {assert (Num> = 0); int factor = 1; int COUNT = 0; int curnumber = 0; int highnumber = 0; int lownumber = 0; while (Num/factor! = 0) {lownumber = num-(Num/factor) * factor; // low digit curnumber = (Num/factor) % 10; // current digit highnumber = (Num/factor)/10; // high digit switch (curnumber) {Case 0: Count + = highnumber * factor; break; Case 1: count + = highnumber * factor + lownumber + 1; break; default: Count + = (highnumber + 1) * factor; break;} factor * = 10;} return count ;}