LeetCode 788. Rotated Digits (旋转数字)

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]

 

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题目标签：String

　　题目给了我们一个N， 让我们找出从 1 到 N 中有几个 good number。Good Number 是每一个digit 都可以旋转，然后旋转后 与 旋转前 是不同的 数字。

　　利用HashMap 把 0，1，8，2，5，6，9 这些可以旋转的数字存入保存。

　　之后利用Map来检查数字的每一个 digit 是否可以旋转，当旋转完之后的新数字 要与 旧数字 不同，才可以算。

　　具体请看code。

 

 

　　

　　

Java Solution:

Runtime beats  23.35% 

完成日期：05/11/2018

关键词：HashMap

关键点：把可以旋转的digit存入map

class Solution 
{
    public int rotatedDigits(int N) 
    {
        int count = 0;
        HashMap<Integer, Integer> map = new HashMap<>();
        
        map.put(0, 0);
        map.put(1, 1);
        map.put(8, 8);
        map.put(2, 5);
        map.put(5, 2);
        map.put(6, 9);
        map.put(9, 6);
        

        for(int i=1; i<=N; i++)
        {
            int number = i;
            int newNumber = 0;
            int m = 1;
            
            while(number > 0)
            {
                int digit = number % 10;     // get the digit

                if(map.containsKey(digit))  // check if digit can be rotated
                {
                    newNumber = map.get(digit) * m + newNumber;
                    number /= 10;
                    m *= 10;
                }
                else    // if digit is invalid
                {
                    break;
                }
            }
            
            if(number == 0 && i != newNumber)
                count++;
                 
        }
        
        return count;
    }
}

参考资料：n/a

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