feat: add solutions to lcof question: 49.Ugly Number

添加《剑指 Offer》题解：面试题49. 丑数

Author: yanglbme

.gitignore

@@ -1,35 +1,3 @@
-/gradle/wrapper/gradle-wrapper.properties
-##----------Android----------
-# build
-*.apk
-*.ap_
-*.dex
-*.class
-bin/
-gen/
-build/
-
-# gradle
-.gradle/
-gradle-app.setting
-!gradle-wrapper.jar
-build/
-
-local.properties
-
-##----------idea----------
-*.iml
 .idea/
-*.ipr
-*.iws
-
-# Android Studio Navigation editor temp files
-.navigation/
-
-##----------Other----------
-# osx
-*~
 .DS_Store
-gradle.properties
-
 .vscode

README.md

@@ -8,8 +8,6 @@
   <a href="https://github.com/doocs/leetcode/issues"><img src="https://badgen.net/github/open-issues/doocs/leetcode" alt="issues"></a>
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 </p>

README_EN.md

@@ -8,8 +8,6 @@
   <a href="https://github.com/doocs/leetcode/issues"><img src="https://badgen.net/github/open-issues/doocs/leetcode" alt="issues"></a>
   <a href="https://github.com/doocs/leetcode/network/members"><img src="https://img.shields.io/github/forks/doocs/leetcode.svg" alt="forks"></a>
   <a href="https://github.com/doocs/leetcode/blob/master/LICENSE"><img src="https://badgen.net/github/license/doocs/leetcode?color=green" alt="LICENSE"></a><br>
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 </p>

lcof/面试题49. 丑数/README.md

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+# [面试题49. 丑数](https://leetcode-cn.com/problems/chou-shu-lcof/)
+
+## 题目描述
+<!-- 这里写题目描述 -->
+我们把只包含因子 2、3 和 5 的数称作丑数（Ugly Number）。求按从小到大的顺序的第 n 个丑数。
+
+**示例:**
+
+```
+输入: n = 10
+输出: 12
+解释: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 是前 10 个丑数。
+```
+
+**说明:**
+
+1. `1` 是丑数。
+2. `n` 不超过 1690。
+
+
+## 解法
+<!-- 这里可写通用的实现逻辑 -->
+
+
+### Python3
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def nthUglyNumber(self, n: int) -> int:
+        if n < 7:
+            return n
+        dp = [1 for _ in range(n)]
+        i2 = i3 = i5 = 0
+        for i in range(1, n):
+            next2, next3, next5 = dp[i2] * 2, dp[i3] * 3, dp[i5] * 5
+            dp[i] = min(next2, next3, next5)
+            if dp[i] == next2:
+                i2 += 1
+            if dp[i] == next3:
+                i3 += 1
+            if dp[i] == next5:
+                i5 += 1
+        return dp[n - 1]
+
+```
+
+### Java
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int nthUglyNumber(int n) {
+        if (n < 7) {
+            return n;
+        }
+        int[] dp = new int[n];
+        dp[0] = 1;
+        int i2 = 0, i3 = 0, i5= 0;
+        for (int i = 1; i < n; ++i) {
+            int next2 = dp[i2] * 2, next3 = dp[i3] * 3, next5 = dp[i5] * 5;
+            dp[i] = Math.min(Math.min(next2, next3), next5);
+            if (dp[i] == next2) {
+                ++i2;
+            }
+            if (dp[i] == next3) {
+                ++i3;
+            }
+            if (dp[i] == next5) {
+                ++i5;
+            }
+        }
+        return dp[n - 1];
+        
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题49. 丑数/Solution.java

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+class Solution {
+    public int nthUglyNumber(int n) {
+        if (n < 7) {
+            return n;
+        }
+        int[] dp = new int[n];
+        dp[0] = 1;
+        int i2 = 0, i3 = 0, i5 = 0;
+        for (int i = 1; i < n; ++i) {
+            int next2 = dp[i2] * 2, next3 = dp[i3] * 3, next5 = dp[i5] * 5;
+            dp[i] = Math.min(Math.min(next2, next3), next5);
+            if (dp[i] == next2) {
+                ++i2;
+            }
+            if (dp[i] == next3) {
+                ++i3;
+            }
+            if (dp[i] == next5) {
+                ++i5;
+            }
+        }
+        return dp[n - 1];
+
+    }
+}

lcof/面试题49. 丑数/Solution.py

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+class Solution:
+    def nthUglyNumber(self, n: int) -> int:
+        if n < 7:
+            return n
+        dp = [1 for _ in range(n)]
+        i2 = i3 = i5 = 0
+        for i in range(1, n):
+            next2, next3, next5 = dp[i2] * 2, dp[i3] * 3, dp[i5] * 5
+            dp[i] = min(next2, next3, next5)
+            if dp[i] == next2:
+                i2 += 1
+            if dp[i] == next3:
+                i3 += 1
+            if dp[i] == next5:
+                i5 += 1
+        return dp[n - 1]