788. Rotated Digits 旋转数字

［抄题］：

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.

 [暴力解法]：

时间分析：

空间分析：

 [优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

［一句话思路］：

至少包括一个2/5/6/9，不能有3/4/7

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

［二刷］：

［三刷］：

［四刷］：

［五刷］：

  [五分钟肉眼debug的结果]：

［总结］：

滥竽充数的一道题

［复杂度］：Time complexity: O(n) Space complexity: O(1)

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

 [代码风格] ：

class Solution {
    public int rotatedDigits(int N) {
        //cc
        if (N == 0) {
            return 0;
        }
        //ini
        int count = 0;
        //for loop
        for (int i = 1; i <= N; i++) {
            if (isValid(i)) {
                count++;
            }
        }
        //return
        return count;
    }
    
    public boolean isValid(int i) {
        boolean trueState = false;
        
        while (i != 0) {
            if (i % 10 == 2) trueState = true;
            if (i % 10 == 5) trueState = true;
            if (i % 10 == 6) trueState = true;
            if (i % 10 == 9) trueState = true;
            if (i % 10 == 3) return false;
            if (i % 10 == 4) return false;
            if (i % 10 == 7) return false;
            i = i / 10;
        }
        
        //return 
        return trueState;
    }
}