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  • Leetcode: 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].
     1 class Solution {
     2     public int rotatedDigits(int N) {
     3         int count = 0;
     4         for (int i = 1; i <= N; i ++) {
     5             if (isValid(i)) count ++;
     6         }
     7         return count;
     8     }
     9     
    10     public boolean isValid(int N) {
    11         /*
    12          Valid if N contains ATLEAST ONE 2, 5, 6, 9
    13          AND NO 3, 4 or 7s
    14          */
    15         boolean validFound = false;
    16         while (N > 0) {
    17             if (N % 10 == 2) validFound = true;
    18             if (N % 10 == 5) validFound = true;
    19             if (N % 10 == 6) validFound = true;
    20             if (N % 10 == 9) validFound = true;
    21             if (N % 10 == 3) return false;
    22             if (N % 10 == 4) return false;
    23             if (N % 10 == 7) return false;
    24             N = N / 10;
    25         }
    26         return validFound;
    27     }
    28 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/11691294.html
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