Ugly Number II -- leetcode

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.

Hint:

1. The naive approach is to call isUgly for every number until you reach the nth one. Most numbers are not ugly. Try to focus your effort on generating only the ugly ones.
2. An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
3. The key is how to maintain the order of the ugly numbers. Try a similar approach of merging from three sorted lists: L1, L2, and L3.
4. Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).

算法一，DP逐個生成ugly numbers

1. ugly number初始爲1

2. 每次循環生成一個新的次大ugly number

即從小到大生成ugly number。

3. 從已經生成的ugly number中，挑選一個，使其*2, 或者*3, *5，得到下一個ugly number。

即 = min(x*2, min(y*3, z*5))  x, y, z爲已經生成的ugly number

class Solution {
public:
    int nthUglyNumber(int n) {
        vector<int> uglys(n, 1);
        int id2 = 0, id3 = 0, id5 = 0;
        for (int i=1; i<n; i++) {
            uglys[i] = min(uglys[id2]*2, min(uglys[id3]*3, uglys[id5]*5));
            if (uglys[i] == uglys[id2]*2)
                ++id2;
            if (uglys[i] == uglys[id3]*3)
                ++id3;
            if (uglys[i] == uglys[id5]*5)
                ++id5;
        }
        return uglys[n-1];
    }
};

算法一開始分配了n個存儲空間，用來存儲n個ugly number。

但在生成新的ugly number時，前面的逐漸開始不需要了。

將vector，改作list，這樣，可以把前面不再需要的ugly number逐漸刪除掉，以節省空間。

class Solution {
public:
    int nthUglyNumber(int n) {
        list<int> uglys;
        uglys.push_back(1);
        auto iter2 = uglys.begin();
        auto iter3 = uglys.begin();
        auto iter5 = uglys.begin();
        for (int i=1; i<n; i++) {
            uglys.push_back(min(*iter2 * 2, min(*iter3 * 3, *iter5 * 5)));
            if (*iter2 * 2 == uglys.back())
                ++iter2;
            if (*iter3 * 3 == uglys.back())
                ++iter3;
            if (*iter5 * 5 == uglys.back())
                ++iter5;
            while (*uglys.begin() < *iter5)
                uglys.pop_front();
        }
        return uglys.back();
    }
};

算法二，顯示歸併

和算法一思路差不多。算法一，隱式有三個隊列，並作歸併。

此處，顯示的設置三個隊列。

1. 每當求得一個新的ugly number，將期*2, *3, *5，並分別送入三個隊列。

2. 對這個三隊列作歸併，求出一個新的ugly number。

3. 重複步驟1，2

class Solution {
public:
    int nthUglyNumber(int n) {
        int ans = 1;
        --n;
        queue<int> L2, L3, L5;
        while (n--) {
            L2.push(ans * 2);
            L3.push(ans * 3);
            L5.push(ans * 5);
            while (L2.front() <= ans) L2.pop();
            while (L3.front() <= ans) L3.pop();
            while (L5.front() <= ans) L5.pop();
            ans = min(L2.front(), min(L3.front(), L5.front()));
        }
        return ans;
    }
};