[LeetCode] Number of Digit One 数字1的个数

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.

Hint:

  1. Beware of overflow.

这道题让我们比给定数小的所有数中1出现的个数,之前有道类似的题 Number of 1 Bits,那道题是求转为二进数后1的个数,博主开始以为这道题也是要用那题的方法,其实不是的,这题实际上相当于一道找规律的题。那么为了找出规律,我们就先来列举下所有含1的数字,并每10个统计下个数,如下所示:

1的个数          含1的数字                                                                        数字范围

1                                                                                                        [1, 9]

11                 0    2  3  4  5  6  7  8  9                              [10, 19]

1                   2                                                                                   [20, 29]

1                   3                                                                                   [30, 39]

1                   4                                                                                   [40, 49]

1                   5                                                                                   [50, 59]

1                   6                                                                                   [60, 69]

1                   7                                                                                   [70, 79]

1                   81                                                                                   [80, 89]

1                   9                                                                                   [90, 99]

11                 00  0  02  03  04  05  06 07  08  09          [100, 109]

21                 0    2  3  4  5  6  7  8  9             [110, 119]

11                 20  2  22  23  24  25  26  27  28  29          [120, 129]

...                  ...                                                                                  ...

通过上面的列举可以发现,100 以内的数字,除了10-19之间有 11 个 ‘1’ 之外,其余都只有1个。如果不考虑 [10, 19] 区间上那多出来的 10 个 ‘1’ 的话,那么在对任意一个两位数,十位数上的数字(加1)就代表1出现的个数,这时候再把多出的 10 个加上即可。比如 56 就有 (5+1)+10=16 个。如何知道是否要加上多出的 10 个呢,就要看十位上的数字是否大于等于2,是的话就要加上多余的 10 个 '1'。那么就可以用 (x+8)/10 来判断一个数是否大于等于2。对于三位数区间 [100, 199] 内的数也是一样,除了 [110, 119] 之间多出的10个数之外,共 21 个 ‘1’,其余的每 10 个数的区间都只有 11 个 ‘1’,所以 [100, 199] 内共有 21 + 11 * 9 = 120 个 ‘1’。那么现在想想 [0, 999] 区间内 ‘1’ 的个数怎么求?根据前面的结果,[0, 99] 内共有 20 个,[100, 199] 内共有 120 个,而其他每 100 个数内 ‘1’ 的个数也应该符合之前的规律,即也是 20 个,那么总共就有 120 + 20 * 9 = 300 个 ‘1’。那么还是可以用相同的方法来判断并累加1的个数,参见代码如下:

解法一:

class Solution {
public:
int countDigitOne(int n) {
int res = , a = , b = ;
while (n > ) {
res += (n + ) / * a + (n % == ) * b;
b += n % * a;
a *= ;
n /= ;
}
return res;
}
};

解法二:

class Solution {
public:
int countDigitOne(int n) {
int res = ;
for (long k = ; k <= n; k *= ) {
long r = n / k, m = n % k;
res += (r + ) / * k + (r % == ? m + : );
}
return res;
}
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/233

类似题目:

Factorial Trailing Zeroes

Digit Count in Range

参考资料:

https://leetcode.com/problems/number-of-digit-one/

https://leetcode.com/problems/number-of-digit-one/discuss/64390/AC-short-Java-solution

https://leetcode.com/problems/number-of-digit-one/discuss/64381/4+-lines-O(log-n)-C++JavaPython

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