【特例】剑指offer——面试题34：丑数

力扣168，
思路链接：https://www.nowcoder.com/questionTerminal/6aa9e04fc3794f68acf8778237ba065b?answerType=1&f=discussion

答案：

Python

class Solution:
    def nthUglyNumber(self, n: int) -> int:
        res = [0] * n
        res[0] = 1
        t2 = t3 = t5 = 0
        for i in range(1, n):
            res[i] = min(res[t2] * 2, min(res[t3] * 3, res[t5] * 5))
            if res[i] == res[t2] * 2:
                t2 += 1
            if res[i] == res[t3] * 3:
                t3 += 1
            if res[i] == res[t5] * 5:
                t5 += 1
        
        return res[-1]

Java

class Solution {
    public int nthUglyNumber(int n) {
        int[] array = new int[n]; 
        array[0] = 1;
        int t2 = 0, t3 = 0, t5 = 0;
        for (int i = 1; i < n; ++i) {
            array[i] = Math.min(array[t2] * 2, Math.min(array[t3] * 3, array[t5] * 5));
            if (array[i] == array[t2] * 2) {
                ++t2;
            }
            if (array[i] == array[t3] * 3) {
                ++t3;
            } 
            if (array[i] == array[t5] * 5) {
                ++t5;
            }
        }
        return array[n - 1];
    }
}

##Solution1:
最容易想到的，也是最不可能AC的

class Solution {
public:
    int GetUglyNumber_Solution(int index) { //输出第index个丑数
        int count = 1;
        if(index == 1)
            return count;
        for(int i = 2; ; i++){
            if(isUglyNum(i) == true){
                count++;
                if(count == index)
                    return i;
            }
        }
    }
    bool isUglyNum(int n) {
        while (n > 1) {
            for (int i = 2; i <= n; i++){
                if (n%i == 0){ //如果n能被i整除，i是因子，对i分小于等于5和大于5两种情况讨论
                    if (i <= 5) {
                        n /= i;
                        break;
                    }
                    if (i > 5)
                        return false;
                }
            }
        }
        return true;
    }
};

##Solution2:
记住这种查数的技巧！！！
答案参考网址：
[1]http://www.cnblogs.com/qqky/p/6964764.html
[2]https://www.nowcoder.com/profile/5810633/codeBookDetail?submissionId=16629921
解题思路：https://blog.youkuaiyun.com/lhanchao/article/details/52079323
本题定义三个变量，对于当前已排好序找到的最大的丑数M，记录恰乘2、乘3、乘5大于当前最大的丑数M的三个位置，然后求三个位置*2 *3 *5的最小值即为下一个丑数的值

class Solution {
public:
   int GetUglyNumber_Solution(int index) {
        if (index < 7) return index;
        vector<int> res(index);
        res[0] = 1;
        int t2 = 0, t3 = 0, t5 = 0, i;
        for (i = 1; i < index; ++i) {
            res[i] = min(res[t2] * 2, min(res[t3] * 3, res[t5] * 5));
            if (res[i] == res[t2] * 2) t2++;
            if (res[i] == res[t3] * 3) t3++;
            if (res[i] == res[t5] * 5) t5++;
        }
        return res[index - 1];
    }
};

再感慨一句，别人的思路真牛逼啊~