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.
fuck!没有看懂题目!!!听说是 一个数 valid 的条件是这个数不包含 3, 4, 7 且包含至少一个 2, 5, 6, 9
最直观的解法:
class Solution(object): def rotatedDigits(self, N): """ :type N: int :rtype: int """ # 一个数 valid 的条件是这个数不包含 3, 4, 7 且包含至少一个 2, 5, 6, 9 def is_valid(n): found = False while n: d = n % 10 if d in {3,4,7}: return False if d in {2,5,6,9}: found = True n /= 10 return found ans = 0 for n in range(1, N+1): if is_valid(n): ans += 1 return ans
看还有dp解法的:
Using a int[] for dp.
dp[i] = 0, invalid number
dp[i] = 1, valid and same number
dp[i] = 2, valid and different number
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;
}
最直观的解法:
class Solution(object): def rotatedDigits(self, N): counts = 0 for num in range(1, N+1): number = str(num) if '3' in number or '7' in number or '4' in number: # This will be an invalid number upon rotation continue # Skip this number and go to next iteration if '2' in number or '5' in number or '6' in number or '9' in number: counts += 1 return counts