# 788. 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]`.

``````class Solution {
public:
int rotatedDigits(int N) {
int res = 0;
for (int i = 1; i <= N; ++i) {
if (check(i)) ++res;
}
return res;
}
bool check(int k) {
string str = to_string(k);
bool flag = false;
for (char c : str) {
if (c == '3' || c == '4' || c == '7') return false;
if (c == '2' || c == '5' || c == '6' || c == '9') flag = true;;
}
return flag;
}
};
``````

``````class Solution {
public:
int rotatedDigits(int N) {
int res = 0;
vector<int> dp(N + 1);
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; ++res;
}
} 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; ++res;
}
}
}
return res;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/788

https://leetcode.com/problems/rotated-digits/

https://leetcode.com/problems/rotated-digits/discuss/117975/Java-dp-solution-9ms

https://leetcode.com/problems/rotated-digits/discuss/264282/Java-O(logN)-0ms-100

https://leetcode.com/problems/rotated-digits/discuss/116530/O(logN)-Time-O(1)-Space

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