[剑指 offer] JT33---Ugly number (The guest officer comes in and draws a lucky number!)

The topic is as follows

Idea and code

Just keep doing it,Either multiply by 2, or multiply by 3, or multiply by 5.
ugly[0]=1;//The first ugly number is 1
ugly[1]=min(1 2,1 3,1 5)=2;// 0 0 0 -->1 0 0
ugly[2] =min(2 2,1 3,1 5)=3;// 1 0 0 -->1 1 0
ugly[3]=min(2 2,2 3,1 5)=4;// 1 1 0 -->2 1 0
ugly[4]=min(3 2,2 3,1 5)=5;// 2 1 0 -->2 1 1
ugly[5]=min(3 2,2 3,2 . 5) =. 6; // 2. 1. 1 -> 2. 3. 1
Ugly [. 6] = min (. 4 2,3 3,2- . 5) =. 8; // 2. 3. 1 -> 2. 4. 1
primary analogy can Calculated all the ugly numbers

class Solution {
    
    
public:
    int GetUglyNumber_Solution(int index) {
    
    
        if(index<=0) return 0;
        vector<int> ugly(index);
        ugly[0]=1;
        int p2=0,p3=0,p5=0;
        for(int i=1;i<index;i++){
    
    
            ugly[i]=min(ugly[p2]*2,min(ugly[p3]*3,ugly[p5]*5));
            if(ugly[i]==ugly[p2]*2) p2++;
            if(ugly[i]==ugly[p3]*3) p3++;
            if(ugly[i]==ugly[p5]*5) p5++;
        }
        return ugly[index-1];
    }
};