[Leetcode] 788. Rotated Digits 解题报告

题目：

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].

思路：

刚开始做的时候以为有什么更好的算法，不用遍历就可以通过公式获得答案，后来发现还是需要一个一个试，所以就没有什么算法方面需要讲的了。

唯一需要注意的是，1,10等不符合题目要求，因为它们rotate之后还是本身，所以我们在检测的过程中就要求数字中至少含有一个2,5,6,9，并且里面不能含有3,4,7。

代码：

class Solution {
public:
    int rotatedDigits(int N) {
        int count = 0;
        for (int i = 2; i <= N; ++i) {
            count += isValid(i);
        }
        return count;
    }
private:
    bool isValid(int N) {
        // Valid if N contains at least one 2, 5, 6, 9, and no 3, 4 or 7s
        bool validFound = false;
        while (N > 0) {
            if (N % 10 == 2) validFound = true;
            if (N % 10 == 5) validFound = true;
            if (N % 10 == 6) validFound = true;
            if (N % 10 == 9) validFound = true;
            if (N % 10 == 3) return false;
            if (N % 10 == 4) return false;
            if (N % 10 == 7) return false;
            N = N / 10;
        }
        return validFound;
    }
};