leetcode--Ugly Number II

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本文提供三种不同的Java实现方案来找到第N个丑数，丑数是指仅包含质因数2、3和5的正整数。文章通过具体代码展示了如何高效地生成丑数序列。

题目：Ugly Number II

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.

One：利用2、3、5之间的乘积值的大小比较。

public class Solution {
    public int nthUglyNumber(int n) {
        int[] uglyNumbers = new int[n];
        uglyNumbers[0] = 1;
        int index2 = 0;
        int index3 = 0;
        int index5 = 0;
        int counter = 1;
        
        while(counter < n){
            int min = minOfThree(uglyNumbers[index2]*2,uglyNumbers[index3]*3,uglyNumbers[index5]*5);
            if(min == uglyNumbers[index2]*2){
                index2++;
            }
            if(min == uglyNumbers[index3]*3){
                index3++;
            }
            if(min == uglyNumbers[index5]*5){
                index5++;
            }
            uglyNumbers[counter] = min;
            counter++;
        }
        return uglyNumbers[n-1];
    }
    private int minOfThree(int x,int y,int z){
        return Math.min(z,Math.min(x,y));
    }
}

Two:进一步简化

public class Solution {
    public int nthUglyNumber(int n) {
        int[] uglyNumbers = new int[n];
        uglyNumbers[0] = 1;
        int index2 = 0;
        int index3 = 0;
        int index5 = 0;
        int counter = 1;
        
        while(counter < n){
            int min = Math.min(uglyNumbers[index5]*5,Math.min(uglyNumbers[index2]*2,uglyNumbers[index3]*3));
            if(min == uglyNumbers[index2]*2)index2++;
            if(min == uglyNumbers[index3]*3)index3++;
            if(min == uglyNumbers[index5]*5)index5++;
            uglyNumbers[counter] = min;
            counter++;
        }
        return uglyNumbers[n-1];
    }
}

Three:利用坐标来操作

public class Solution {
    public int nthUglyNumber(int n) {
        int[] uglyNumbers = new int[n];
        uglyNumbers[0] = 1;
        int index2 = 0;
        int index3 = 0;
        int index5 = 0;
        int p = 1;
        
        while(p < n){
            uglyNumbers[p] = Math.min(uglyNumbers[index5]*5,Math.min(uglyNumbers[index2]*2,uglyNumbers[index3]*3));
            if(uglyNumbers[p] >= uglyNumbers[index2]*2)index2++;
            if(uglyNumbers[p] >= uglyNumbers[index3]*3)index3++;
            if(uglyNumbers[p] >= uglyNumbers[index5]*5)index5++;
            p++;
        }
        return uglyNumbers[n-1];
    }
}