LeetCode #1088. Confusing Number II

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探讨一种特殊数字——混淆数字，这类数字在旋转180度后会变成一个不同的有效数字。本文通过示例解释了如何识别1到N之间的混淆数字，并提供了一个算法解决方案。

题目描述：

We can rotate digits by 180 degrees to form new digits. When 0, 1, 6, 8, 9 are rotated 180 degrees, they become 0, 1, 9, 8, 6 respectively. When 2, 3, 4, 5 and 7 are rotated 180 degrees, they become invalid.

A confusing number is a number that when rotated 180 degrees becomes a different number with each digit valid.(Note that the rotated number can be greater than the original number.)

Given a positive integer N, return the number of confusing numbers between 1 and N inclusive.

Example 1:

Input: 20
Output: 6
Explanation: 
The confusing numbers are [6,9,10,16,18,19].
6 converts to 9.
9 converts to 6.
10 converts to 01 which is just 1.
16 converts to 91.
18 converts to 81.
19 converts to 61.

Example 2:

Input: 100
Output: 19
Explanation: 
The confusing numbers are [6,9,10,16,18,19,60,61,66,68,80,81,86,89,90,91,98,99,100].

Note:

1 <= N <= 10^9

class Solution {
public:
    int confusingNumberII(int N) {
        int count=0;
        search(0,N,count);
        return count;
    }
    
    // 旋转之后可能溢出，要用long long存储
    void search(long long cur, int N, int& count)
    {
        if(cur>N) return;
        if(cur!=rotate(cur)) count++;
        if(cur>0) search(cur*10,N,count); // 必须大于零，否则无限循环
        search(cur*10+1,N,count);
        search(cur*10+6,N,count);
        search(cur*10+8,N,count);
        search(cur*10+9,N,count);
    }
    
    long long rotate(long long n)
    {
        long long m=0;
        while(n>0) 
        {
            int x=n%10;
            if(x==6) x=9;
            else if(x==9) x=6;
            m*=10;
            m+=x;
            n/=10;
        }
        return m;
    }
};