From cbdaa6448d8e4f85e54466e510fe828cac16728c Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Tue, 11 Feb 2020 18:25:03 +0800
Subject: [PATCH] update 264

---
 solutions/264. Ugly Number II.md | 20 +++++++++-----------
 1 file changed, 9 insertions(+), 11 deletions(-)

diff --git a/solutions/264. Ugly Number II.md b/solutions/264. Ugly Number II.md
index 922fa9e..e36b301 100644
--- a/solutions/264. Ugly Number II.md	
+++ b/solutions/264. Ugly Number II.md	
@@ -6,6 +6,8 @@
 
 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. 
 
+时空复杂度均为O(n)
+
 # C++
 ``` C++
 class Solution {
@@ -14,18 +16,14 @@ public:
         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++;
+        int tmp2 = 2, tmp3 = 3, tmp5 = 5, next;
+        while(--n){
+            next = min(tmp2, min(tmp3, tmp5));
             ugly.push_back(next);
+
+            if(next == tmp2) tmp2 = ugly[++idx2] * 2;
+            if(next == tmp3) tmp3 = ugly[++idx3] * 3;
+            if(next == tmp5) tmp5 = ugly[++idx5] * 5;
         }
         return ugly.back();
     }