leetcode 264. Ugly Number II

Write a program to find the 

n

-th ugly number.

Ugly numbers are positive numbers whose prime factors only include 

2, 3, 5

. For example, 

1,
2, 3, 4, 5, 6, 8, 9, 10, 12

 is the sequence of the first 

10

 ugly numbers.

Note that 

1

 is typically treated as an ugly number.

class Solution {
struct Node
{
int val;
vector<int>bb;
Node(int x) :val(x){
bb.push_back(5);
bb.push_back(3);
bb.push_back(2);
}
};
public:
int nthUglyNumber(int n) {
clock_t start, finish;
double Total_time;
start = clock();
if (n == 1)
return 1;
vector<int>cc;
cc.push_back(1);
vector<Node>candi;
candi.push_back(Node(1));
while (cc.size() != n)
{
int min = 2147483647;
int ii = -1;
int i = 0;
while (candi.size() > i)
{
while (candi[i].val*candi[i].bb.back() <= cc.back())
candi[i].bb.pop_back();
if (candi[i].val*candi[i].bb.back() < min)
{
min = candi[i].val*candi[i].bb.back();
ii = i;
}
i++;
if (i < candi.size() && candi[i].bb.size() == 3 && candi[i].val * 2 >= min)
break;
}
cc.push_back(min);
candi.push_back(Node(min));
candi[ii].bb.pop_back();
if (candi[ii].bb.empty())
candi.erase(candi.begin() + ii, candi.begin() + ii + 1);

}
finish = clock();
Total_time = (double)(finish - start) / CLOCKS_PER_SEC;
printf("%f seconds\n", Total_time);
return cc.back();
}
};

这个超时了，想了想性能的瓶颈可能是每次找min时计算太多，其实上次算出来的结果可以存下来，这样可能会减少一些计算。

改进之后accept

class Solution {
public:
int nthUglyNumber(int n) {
clock_t start, finish;
double Total_time;
start = clock();
if (n == 1)
return 1;
vector<int>cc;
cc.push_back(1);
set<int>m2, m3, m5;
m2.insert(1);
m3.insert(1);
m5.insert(1);
while (cc.size() != n)
{
while (*m2.begin() * 2 <= cc.back())
m2.erase(m2.begin());
int a2 = *m2.begin() * 2;
while (*m3.begin() * 3 <= cc.back())
m3.erase(m3.begin());
int a3 = *m3.begin() * 3;
while (*m5.begin() * 5 <= cc.back())
m5.erase(m5.begin());
int a5 = *m5.begin() * 5;
int min = a2 < a3 ? a2 : a3;
min = min < a5 ? min : a5;
if (min == a2)
m2.erase(m2.begin());
if (min==a3)
m3.erase(m3.begin());
if (min==a5)
m5.erase(m5.begin());
m2.insert(min);
m3.insert(min);
m5.insert(min);
cc.push_back(min);
}

finish = clock();
Total_time = (double)(finish - start) / CLOCKS_PER_SEC;
printf("%f seconds\n", Total_time);
return cc.back();
}
};