[剑指offer]面试题第[49]题[Leetcode][第264题][JAVA][丑数][动态规划][堆]

灰太狼 2023-02-27 13:56 12阅读 0赞

#### 【问题描述】\[中等\] ####

#### 【解答思路】 ####

###### 1. 动态规划 ######

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70]

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 1]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 2]

**时间复杂度：O(N) 空间复杂度：O(N)**  
![在这里插入图片描述][20200718122948967.png]

class Solution { 
        public int nthUglyNumber(int n) { 
            int a = 0, b = 0, c = 0;
            int[] dp = new int[n];
            dp[0] = 1;
            for(int i = 1; i < n; i++) { 
                int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5;
                dp[i] = Math.min(Math.min(n2, n3), n5);
                if(dp[i] == n2) a++;
                if(dp[i] == n3) b++;
                if(dp[i] == n5) c++;
            }
            return dp[n - 1];
        }
    }
    
    作者：jyd
    链接：https://leetcode-cn.com/problems/chou-shu-lcof/solution/mian-shi-ti-49-chou-shu-dong-tai-gui-hua-qing-xi-t/

###### 2. 堆 ######

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 3]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 4]

**时间复杂度：O(N) 空间复杂度：O(1)**

class Solution { 
        public int nthUglyNumber(int n) { 
            PriorityQueue<Long> pq = new PriorityQueue<>();
            Set<Long> s = new HashSet<>();
            //初始化,放进堆和set，发现1要开Long数组才可以
            long[] primes = new long[]{ 2, 3, 5};
            for (long prime : primes) { 
                pq.offer(prime);
                s.add(prime);
            }
            long num = 1;
            for (int i = 1; i < n; i++) { 
                num = pq.poll();
                //遍历三个因子
                for (int j = 0; j < 3; j++) { 
                    if (!s.contains(num * primes[j])) { 
                        pq.offer(num * primes[j]);
                        s.add(num * primes[j]);
                    }
                }
            }
            return (int) num;
        }
    }
    
    作者：jerry_nju
    链接：https://leetcode-cn.com/problems/chou-shu-lcof/solution/dui-he-dong-tai-gui-hua-si-lu-xiang-jie-by-jerry_n/

#### 【总结】 ####

###### 1.动态规划流程 ######

第 1 步：设计状态  
第 2 步：状态转移方程  
第 3 步：考虑初始化  
第 4 步：考虑输出  
第 5 步：考虑是否可以状态压缩

###### 2.优先队列 按大小弹出 溢出问题long 去重set ######

参考链接：  
https://leetcode-cn.com/problems/chou-shu-lcof/solution/dui-he-dong-tai-gui-hua-si-lu-xiang-jie-by-jerry\_n/  
参考链接：https://leetcode-cn.com/problems/chou-shu-lcof/solution/mian-shi-ti-49-chou-shu-dong-tai-gui-hua-qing-xi-t/

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[20200718122948967.png]: https://img-blog.csdnimg.cn/20200718122948967.png
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