788. Rotated Digits

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Problem

An integer x is a good 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. For example:

• 0, 1, and 8 rotate to themselves,
• 2 and 5 rotate to each other (in this case they are rotated in a different direction, in other words, 2 or 5 gets mirrored),
• 6 and 9 rotate to each other, and
• the rest of the numbers do not rotate to any other number and become invalid.

Given an integer n, return the number of good integers in the range [1, n].

Example 1:

Input: n = 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.

Example 2:

Input: n = 1
Output: 0

Example 3:

Input: n = 2
Output: 1

Constraints:

• $1 <= n <= 10^4$

Code

/**
 * dp[i] = 0, invalid number
 * dp[i] = 1, valid and same number
 * dp[i] = 2, valid and different number
 */

class Solution {
    public int rotatedDigits(int N) {
        int[] dp = new int[N + 1];
        int count = 0;
        for (int i = 0; i <= N; i++) {
            if (i < 10) {
                if (i == 0 || i == 1 || i == 8) {
                    dp[i] = 1;
                } else if (i == 2 || i == 5 || i == 6 || i == 9) {
                    dp[i] = 2;
                    count++;
                }
            } else {
                int a = dp[i / 10], b = dp[i % 10];
                if (a == 1 && b == 1) {
                    dp[i] = 1;
                } else if (a >= 1 && b >= 1) {
                    dp[i] = 2;
                    count++;
                }
            }
        }

        return count;
    }
}