LeetCode之Ugly Number II

/*根据提示，保存一个有序的丑数数组，然后通过不断乘以2,3,5，
从中选出最小的数作为下一个丑数。*/
class Solution {
public:
int nthUglyNumber(int n) {
vector<int> uglies(1, 1);
//uglies.reserve(n);
int i2(0), i3(0), i5(0);
while(uglies.size() < n){
int p2 = 2 * uglies[i2], p3 = 3 * uglies[i3], p5 = 5 * uglies[i5];
int res = min(p2, min(p3, p5));
if(res == p2) ++i2;
if(res == p3) ++i3;
if(res == p5) ++i5;
uglies.push_back(res);
}
return uglies.back();
}
};