Enter an integer n and find the number of occurrences of 1 in the decimal representation of n integers from 1 to n.
For example, if you enter 12, the numbers containing 1 in the integers from 1 to 12 are 1, 10, 11, and 12, and 1 appears 5 times in total.
Example 1:
输入：n = 12
输出：5
Example 2:
输入：n = 13
输出：6
Restrictions:
1 <= n <2^31
Note: This question is the same as the 233 question on the main website: https://leetcode-cn.com/problems/number-of-digit-one/
Problem-solving ideas:
How to calculate the number of occurrences of 1 in a certain position:
According to the different cur value of the current bit, it can be divided into the following three situations:
1. When cur = 0 : The number of occurrences of this bit 1 is only determined by the high bit high. The calculation formula is:
high×digit
as shown in the figure below. Taking n = 2304 as an example, find the 1 of digit = 10 (that is, the tens digit) The number of occurrences.
2. When cur = 1 : The number of occurrences of this bit 1 is determined by the high bit and the low bit low. The calculation formula is:
high×digit+low+1
as shown in the figure below. Taking n = 2314 as an example, find digit = 10 ( That is, the number of occurrences of 1 in the tens place.
3. When cur = 2, 3,...,9 : The number of occurrences of this bit 1 is only determined by the high bit high. The calculation formula is:
(high+1)×digit
As shown in the figure below, taking n = 2324 as an example, find the number of 1 occurrences of digit = 10 (that is, the tens place).
class Solution {
public:
int countDigitOne(int n) {
long digit = 1;
int res = 0;
int high = n / 10, cur = n % 10, low = 0;//cur是n除以10的余数，后面的cur是high的余数
while(high != 0 || cur != 0) {
// 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出
if(cur == 0) res += high * digit;
else if(cur == 1) res += high * digit + low + 1;
else res += (high + 1) * digit;//
low += cur * digit;//low就是cur一轮一轮叠加得到；此处参数是用作下一轮
cur = high % 10;//cur是high的余数；；下一轮用
high /= 10;//high从右边减少一位；下一轮用
digit *= 10;//digit乘以10；下一轮用
}
return res;
}
};
Complexity analysis:
Time complexity O(log n): The calculation operation in the loop uses O(1) time; the number of loops is the number of digits of the number n, that is, log 10 n, so the loop uses O(log n) time.
Space complexity O(1): Several variables use extra space of constant size.
Author: jyd
link: https: //leetcode-cn.com/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-lcof/solution/mian-shi-ti-43-1n-zheng-shu -zhong-1-chu-xian-de-2/
Source: LeetCode (LeetCode)
copyright belongs to the author. For commercial reprints, please contact the author for authorization, and for non-commercial reprints, please indicate the source.