LeetCode|Rotated Digits

最新推荐文章于 2021-06-07 10:58:37 发布

DouMiaoO_Oo

最新推荐文章于 2021-06-07 10:58:37 发布

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本文探讨了一个有趣的数学问题：旋转数字。定义一个好数字为在每个数字分别旋转180度后得到的有效且与原数不同的数字。文章详细介绍了哪些数字在旋转后仍保持有效，如0、1和8旋转后不变，2和5、6和9互转，而其它数字则变为无效。通过一个示例说明了如何计算从1到N范围内的好数字数量。

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Rotated Digits

X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X. Each digit must be rotated - we cannot choose to leave it alone.

A number is valid if each digit remains a digit after rotation. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each other; 6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number and become invalid.

Now given a positive number N, how many numbers X from 1 to N are good?

Example:
Input: 10
Output: 4
Explanation: 
There are four good numbers in the range [1, 10] : 2, 5, 6, 9.
Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.

Note:

N will be in range [1, 10000].

class Solution {
public:
    int rotatedDigits(int N) {
        // 0 1 8, 2 5, 6 9 
        int cnt = 0;
        for(int i = 1; i <= N; ++i){
            cnt += valid(i);
        } return cnt;
    }
private:
    int valid(int x){
        int appear = 0;
        while(x){
            int bit = x%10;
            if(bit == 2 || bit == 5 || bit == 6 || bit == 9){
                appear = 1;
            } else if(bit == 3 || bit == 4 || bit == 7){
                return 0;
            } x /= 10;
        }
        return appear;
    }
};