From b187e57d054c4a7d526d5232b87120b6ebfc7292 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Tue, 29 Oct 2019 19:59:48 +0800
Subject: [PATCH] Create 264. Ugly Number II.md

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 solutions/264. Ugly Number II.md | 32 ++++++++++++++++++++++++++++++++
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 create mode 100644 solutions/264. Ugly Number II.md

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+# 思路
+要求第n个ugly数. 由于ugly集合相对于整个int集合是很小的, 所以此题用蛮力法不断判断某个数是否ugly肯定会超时. 所以我们应该从直接构造ugly数着手.
+
+那么如何直接构造ugly数呢? 根据ugly数的定义可知, 如果x, y, z都是ugly的, 那么2x, 3y, 5z都是ugly数, 所以我们可以从初始x=y=z=1开始不断构造ugly数.
+
+题目要求第n大的ugly数, 所以可以维护一个从小到大的ugly数组, 每次取2x, 3y, 5z三者最小的push进数组作为下一个ugly数, 详情可见代码. 
+
+# C++
+``` C++
+class Solution {
+public:
+    int nthUglyNumber(int n) {
+        vector<int>ugly{1};
+        
+        int idx2 = 0, idx3 = 0, idx5 = 0;
+        int tmp2, tmp3, tmp5;
+        while(1){
+            if(!(--n)) break; 
+            tmp2 = ugly[idx2] * 2;
+            tmp3 = ugly[idx3] * 3;
+            tmp5 = ugly[idx5] * 5;
+            
+            int next = min(tmp2, min(tmp3, tmp5));
+            if(next == tmp2) idx2++;
+            if(next == tmp3) idx3++;
+            if(next == tmp5) idx5++;
+            ugly.push_back(next);
+        }
+        return ugly.back();
+    }
+};
+```