面试题49. 丑数

一时失言乱红尘 2023-02-18 03:10 10阅读 0赞

# 题目： #

[面试题49. 丑数][49.]  
[264. 丑数 II][264. _ II]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5eF9seQ_size_16_color_FFFFFF_t_70]

# 题解：动态规划(三指针) #

## 1. 解释一： ##

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5eF9seQ_size_16_color_FFFFFF_t_70 1]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5eF9seQ_size_16_color_FFFFFF_t_70 2]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5eF9seQ_size_16_color_FFFFFF_t_70 3]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5eF9seQ_size_16_color_FFFFFF_t_70 4]

## 2. 解释二： ##

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5eF9seQ_size_16_color_FFFFFF_t_70 5]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5eF9seQ_size_16_color_FFFFFF_t_70 6]

## 3. 解释三： ##

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5eF9seQ_size_16_color_FFFFFF_t_70 7]

## 4. 解释四： ##

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5eF9seQ_size_16_color_FFFFFF_t_70 8]

# 代码：动态规划(三指针) #

public class 面试题49 { 
    
        public static int nthUglyNumber(int n) { 
            int a = 0; // 索引 a 指向 0
            int b = 0; // 索引 b 指向 0
            int c = 0; // 索引 c 指向 0
            int dp[] = new int[n]; // dp[i] 代表第 i + 1 个丑数
            dp[0] = 1; // 第一个丑数为 1
            for(int i = 1; i < n; i++)
            { 
                int n2 = dp[a] * 2;
                int n3 = dp[b] * 3;
                int n5 = dp[c] * 5;
                dp[i] = Math.min(Math.min(n2, n3), n5);
                if(dp[i] == n2)
                { 
                    a++; // 索引 a 向后移动一个元素
                }
                if(dp[i] == n3)
                { 
                    b++; // 索引 b 向后移动一个元素
                }
                if(dp[i] == n5)
                { 
                    c++; // 索引 c 向后移动一个元素
                }
            }
            return dp[n - 1]; // 返回第 n 个丑数
        }
    
        public static void main(String[] args) { 
            int n = 10;
            int res = nthUglyNumber(n);
            System.out.println(res);
        }
    }

# 参考： #

1.  [面试题49. 丑数（动态规划，清晰图解）][49. 1]
2.  [丑数II， 合并 3 个有序数组， 清晰的推导思路][II_ _ 3 _]
3.  [丑数 II][II]
4.  [三指针方法的理解方式][Link 1]

[49.]: https://leetcode-cn.com/problems/chou-shu-lcof/
[264. _ II]: https://leetcode-cn.com/problems/ugly-number-ii/
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[49. 1]: https://leetcode-cn.com/problems/chou-shu-lcof/solution/mian-shi-ti-49-chou-shu-dong-tai-gui-hua-qing-xi-t/
[II_ _ 3 _]: https://leetcode-cn.com/problems/chou-shu-lcof/solution/chou-shu-ii-qing-xi-de-tui-dao-si-lu-by-mrsate/
[II]: https://leetcode-cn.com/problems/ugly-number-ii/solution/chou-shu-ii-by-leetcode/
[Link 1]: https://leetcode-cn.com/problems/ugly-number-ii/solution/san-zhi-zhen-fang-fa-de-li-jie-fang-shi-by-zzxn/