264. 丑数 II

 1 //动态方程dp[i]=min(dp[p_2]*2,dp[p_3]*3,dp[p_5]*5)
 2 class Solution
 3 {
 4 public:
 5     int nthUglyNumber(int n)
 6     {
 7         vector<int> dp(n);
 8         dp[0] = 1;
 9         int p2,p3,p5;
10         p2 = p3 = p5 = 0;
11         for (int i = 1;i < n;++ i)
12         {
13             dp[i] = min(2 * dp[p2],min(3 * dp[p3],5 * dp[p5]));
14             if(dp[i] == 2 * dp[p2]) ++p2;
15             if(dp[i] == 3 * dp[p3]) ++p3;
16             if(dp[i] == 5 * dp[p5]) ++p5;
17         }
18         return dp.at(n-1);
19     }
20 };

原文地址：https://www.cnblogs.com/yuhong1103/p/12700284.html