牛客OJ：丑数

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方法一，类似筛法素数，只能通过84%的数据。

const int maxn = 10000001;
bool num[maxn];

class Solution {
public:
    int GetUglyNumber_Solution(int index) {
        if(index <= 0) return 0;
        memset(num,0,sizeof(num));
        num[1] = true;
        int cnt = 0;
        for(int i=0;i<maxn;i++){
            if(num[i]){
                cnt ++;
                if(cnt == index) return i;
                num[min(i*2,maxn)] = true;
                num[min(i*3,maxn)] = true;
                num[min(i*5,maxn)] = true;
            }
        }
        return 0;
    }
};

方法二，找M2, M3, M5：

class Solution {
public:
    
int GetUglyNumber_Solution(int index) {
    if(index <= 0) return 0;
    vector<int> ans;
    ans.push_back(1);
    int loc2 = 0;
    int loc3 = 0;
    int loc5 = 0;
    for(int i=2;i<=index;i++){
        ans.push_back(min(ans[loc5]*5,min(ans[loc3]*3,ans[loc2]*2)));
        while(ans[loc2]*2 <= ans[i-1]) loc2++;
        while(ans[loc3]*3 <= ans[i-1]) loc3++;
        while(ans[loc5]*5 <= ans[i-1]) loc5++;
    }
    return ans[ans.size()-1];
}
};