https://leetcode.com/problems/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]
.
代码:
class Solution { public: int rotatedDigits(int N) { int ans = 0; for(int i = 1; i <= N; i ++) { if(isgood(i)) ans ++; } return ans; } bool isgood(int x) { string s = to_string(x); bool flag = false; for(int i = 0; i < s.length(); i ++) { if(s[i] == '3' || s[i] == '7' || s[i] == '4') return false; if(s[i] == '2' || s[i] == '5' || s[i] == '6' || s[i] == '9') flag = true; } return flag; } };
如果数字里出现 3 4 7 数字的话就不是好数字 逐位判断 还是 emmmm 阔以的吧