剑指offer-题34：丑数

题目描述

把只包含因子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。

实验平台：牛客网

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解决思路：

java:

public class Solution {
    public int GetUglyNumber_Solution(int index) {
        if (index <= 0) {
            return 0;
        } else {
            int[] uglyNumarray = new int[index];
            uglyNumarray[0] = 1;
            int uglyNumIndex = 1;
            int multiply2Index = 0;
            int multiply3Index = 0;
            int multiply5Index = 0;

            while (uglyNumIndex < index) {
                int min = min(uglyNumarray[multiply2Index] * 2, uglyNumarray[multiply3Index] * 3, uglyNumarray[multiply5Index] * 5);
                uglyNumarray[uglyNumIndex] = min;

                while (uglyNumarray[multiply2Index] * 2 <= uglyNumarray[uglyNumIndex]) {
                    multiply2Index++;
                }
                while (uglyNumarray[multiply3Index] * 3 <= uglyNumarray[uglyNumIndex]) {
                    multiply3Index++;
                }
                while (uglyNumarray[multiply5Index] * 5 <= uglyNumarray[uglyNumIndex]) {
                    multiply5Index++;
                }
                uglyNumIndex++;
            }
            return uglyNumarray[uglyNumIndex - 1];
        }
    }

    public int min(int a, int b, int c) {
        int min = a < b ? a : b;
        min = min < c ? min : c;
        return min;
    }
}

python:

class Solution:
    def GetUglyNumber_Solution(self, index):
        # write code here
        if index <= 0:
            return 0
        else:
            ugly_num_list = [1]
            ugly_num_index = 1
            multiply2index = 0
            multiply3index = 0
            multiply5index = 0
            while ugly_num_index < index:
                min = self.min(ugly_num_list[multiply2index] * 2, ugly_num_list[multiply3index] * 3, ugly_num_list[multiply5index] * 5)
                ugly_num_list.append(min)
                while ugly_num_list[multiply2index] * 2 <= ugly_num_list[ugly_num_index]:
                    multiply2index += 1
                while ugly_num_list[multiply3index] * 3 <= ugly_num_list[ugly_num_index]:
                    multiply3index += 1
                while ugly_num_list[multiply5index] * 5 <= ugly_num_list[ugly_num_index]:
                    multiply5index += 1
                ugly_num_index += 1
            return ugly_num_list[ugly_num_index - 1]

    def min(self, a, b, c):
        min = (a if a < b else b)
        return min if min < c else c