788. Rotated Digits

788. Rotated Digits

  • 788. Rotated Digits
    • 题目
    • 解决


题目

Leetcode题目

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].

解决

  • 时间复杂度: O(n * n)
  • 空间复杂度:O(n)
class Solution {
public:
    int rotatedDigits(int N) {
        int result = 0;
        vector<int> good = {2, 5, 6, 9};
        vector<int> bad = {3, 4, 7};
        vector<int>::iterator g, b;
        for (int i = 1; i <= N; i++) {
            bool flag = false;
            int num = i;
            while (num) {
                int last = num % 10;
                g = find(good.begin(), good.end(), last);
                b = find(bad.begin(), bad.end(), last);
                if (b != bad.end()) {
                    flag = false;
                    break;
                }
                if (g != good.end()) flag = true;
                num /= 10;
            }
            if (flag) result++;
        }
        return result;
    }
};

你可能感兴趣的:(leetcode,算法)