From 07d219ecd1fb75337d6d5722a1b63635e4f19057 Mon Sep 17 00:00:00 2001
From: Yufan Sheng <18829237653@163.com>
Date: Fri, 4 Oct 2024 21:05:14 +1000
Subject: [PATCH 1/2] =?UTF-8?q?=E6=9B=B4=E6=AD=A3=E5=A4=8D=E6=9D=82?=
 =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90=E4=B8=AD=E7=9A=84=E7=AC=94=E8=AF=AF?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0452.用最少数量的箭引爆气球.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md
index 318c3035..10fe5771 100644
--- a/problems/0452.用最少数量的箭引爆气球.md
+++ b/problems/0452.用最少数量的箭引爆气球.md
@@ -110,7 +110,7 @@ public:
 ```
 
 * 时间复杂度：O(nlog n)，因为有一个快排
-* 空间复杂度：O(1)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间
+* 空间复杂度：O(n)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间
 
 可以看出代码并不复杂。
 

From 2b7fc7d1e183ebcce3930b0235f7f646a6ab3871 Mon Sep 17 00:00:00 2001
From: Yufan Sheng <18829237653@163.com>
Date: Fri, 4 Oct 2024 21:22:50 +1000
Subject: [PATCH 2/2] =?UTF-8?q?=E5=88=A0=E9=99=A4=E5=86=97=E4=BD=99?=
 =?UTF-8?q?=E4=BB=A3=E7=A0=81=EF=BC=8C=E5=A2=9E=E5=8A=A0=E8=AF=A6=E7=BB=86?=
 =?UTF-8?q?=E6=B3=A8=E9=87=8A?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 .../0452.用最少数量的箭引爆气球.md | 22 ++++++++++++-------
 1 file changed, 14 insertions(+), 8 deletions(-)

diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md
index 10fe5771..2099275d 100644
--- a/problems/0452.用最少数量的箭引爆气球.md
+++ b/problems/0452.用最少数量的箭引爆气球.md
@@ -180,19 +180,25 @@ class Solution:
 ```python
 class Solution: # 不改变原数组
     def findMinArrowShots(self, points: List[List[int]]) -> int:
+        if len(points) == 0:
+            return 0
+            
         points.sort(key = lambda x: x[0])
-        sl,sr = points[0][0],points[0][1]
+
+        # points已经按照第一个坐标正序排列，因此只需要设置一个变量，记录右侧坐标（阈值）
+        # 考虑一个气球范围包含两个不相交气球的情况：气球1: [1, 10], 气球2: [2, 5], 气球3: [6, 10]
+        curr_min_right = points[0][1]
         count = 1
+        
         for i in points:
-            if i[0]>sr:
-                count+=1
-                sl,sr = i[0],i[1]
+            if i[0] > curr_min_right:
+                # 当气球左侧大于这个阈值，那么一定就需要在发射一只箭，并且将阈值更新为当前气球的右侧
+                count += 1
+                curr_min_right = i[1]
             else:
-                sl = max(sl,i[0])
-                sr = min(sr,i[1])
+                # 否则的话，我们只需要求阈值和当前气球的右侧的较小值来更新阈值
+                curr_min_right = min(curr_min_right, i[1])
         return count
-
-
 ```
 ### Go
 ```go