丑数

题目描述

把只包含质因子2、3和5的数称作丑数（Ugly Number）。例如6、8都是丑数，但14不是，因为它包含质因子7。 习惯上我们把1当做是第一个丑数。求按从小到大的顺序的第N个丑数。

示例1

输入

7

返回值

8

使用常规思路会导致超时：

class Solution {
public:
    int GetUglyNumber_Solution(int index) {
        if(index<=0) return 0;
        
        int count = 1;
        int i = 1;
        while(count<index){
            i++;
            if(isUglyNum(i)){
                count++;
            }
        }
        return i;
    }
    bool isUglyNum(int n){
        while(n != 1 && n%2 == 0){
            n /= 2;
        }
        while(n != 1 && n%3 == 0){
            n /= 3;
        }
        while(n != 1&& n%5 == 0){
            n /= 5;
        }
        if(n == 1)
            return true;
        else
            return false;
    }
};

利用三个队列进行存储：

class Solution {
public:
    int GetUglyNumber_Solution(int index) {
        if(index<=0) return 0;
        queue<int> queue2;
        queue<int> queue3;
        queue<int> queue5;
        vector<int> array;
        array.push_back(1);

        int i = 0;

        while(i+1<index){
            queue2.push(2*array[i]);
            queue3.push(3*array[i]);
            queue5.push(5*array[i]);
            if(queue2.front()<=queue3.front() && queue2.front()<=queue5.front()){
                array.push_back(queue2.front());
                if(queue2.front()==queue3.front())
                    queue3.pop();
                if(queue2.front()==queue5.front())
                    queue5.pop();
                queue2.pop();
            }
            else if(queue3.front()<=queue2.front() && queue3.front()<=queue5.front()){
                array.push_back(queue3.front());
                if(queue3.front()==queue2.front())
                    queue2.pop();
                if(queue3.front()==queue5.front())
                    queue5.pop();
                queue3.pop();
            }
            else{
                array.push_back(queue5.front());
                if(queue5.front()==queue2.front())
                    queue2.pop();
                if(queue5.front()==queue3.front())
                    queue3.pop();
                queue5.pop();
            }
            i++;
        }
        int temp = array[i];
        array.clear();
        queue2.empty();
        queue3.empty();
        queue5.empty();
        return array[i];
    }
};

 使用一个数组和三个指针来实现：

class Solution {
public:
    int GetUglyNumber_Solution(int index) {
        if(index<=0) return 0;
        int * array = new int[3*index+1];
        array[0] = 1;
        int i = 0; //记录位置
        int count = 1; //记录个数
        int index1 = 1;
        int index2 = 2;
        int index3 = 3;
        int j = 1;//记录下一个要插入数据的位置
        while(count<index){
            array[j++] = array[i] * 2;
            array[j++] = array[i] * 3;
            array[j++] = array[i] * 5;
            if(array[index1]<=array[index2] && array[index1]<=array[index3]){
                i = index1;
                if(array[index1]==array[index2])
                    index2 += 3;
                if(array[index1]==array[index3])
                    index3 += 3;
                index1 += 3;
            }else if(array[index2]<=array[index1] && array[index2]<=array[index3]){
                i = index2;
                if(array[index2]==array[index3])
                    index3 += 3;
                index2 += 3;
            }else{
                i = index3;
                index3 += 3;
            }
            count++;
        }
        int result = array[i];
        delete array;
        return result;
    }
};

使用空间更少的方式进行计算

public class Solution {
    public int GetUglyNumber_Solution(int index) {
        if(index<=0) return 0;
        int[] array = new int[index];
        array[0] = 1;
        int nextIndex = 1;
        int index2=0;
        int index3=0;
        int index5=0;
        int min;
        while(nextIndex < index){
            min = minNum(array[index2]*2,array[index3]*3,array[index5]*5);
            array[nextIndex++] = min;
            if(min == array[index2]*2)
                index2++;
            if(min == array[index3]*3)
                index3++;
            if(min == array[index5]*5)
                index5++;
        }
        return array[nextIndex-1];
    }
    int minNum(int a,int b,int c){
        int min = (a < b ? a:b);
        min = min < c ? min : c;
        return min;
    }
}