标签:possible ++i out lower clu turn mem inpu bec
Problem:
A self-dividing number is a number that is divisible by every digit it contains.
For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.
Also, a self-dividing number is not allowed to contain the digit zero.
Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.
Example 1:
Input:
left = 1, right = 22
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]
Note:
思路:
Solution (C++):
vector<int> selfDividingNumbers(int left, int right) {
vector<int> res;
for (int i = left; i <= right; ++i) {
if (isSelfDividing(i)) res.push_back(i);
}
return res;
}
bool isSelfDividing(int n) {
if (n == 0) return false;
if (n < 10) return true;
vector<int> vec;
int tmp = n;
while (n) {
if (n%10 == 0) return false;
vec.push_back(n%10);
n /= 10;
}
for (auto v : vec) {
if (tmp % v) return false;
}
return true;
}
性能:
Runtime: 40 ms??Memory Usage: 10.6 MB
思路:
Solution (C++):
性能:
Runtime: ms??Memory Usage: MB
标签:possible ++i out lower clu turn mem inpu bec
原文地址:https://www.cnblogs.com/dysjtu1995/p/12729549.html