From 8efc21f92d6485b52d7d24391da9d88090c5bd59 Mon Sep 17 00:00:00 2001
From: LiangDazhu <42199191+LiangDazhu@users.noreply.github.com>
Date: Wed, 19 May 2021 23:23:38 +0800
Subject: [PATCH] =?UTF-8?q?Update=200452.=E7=94=A8=E6=9C=80=E5=B0=91?=
=?UTF-8?q?=E6=95=B0=E9=87=8F=E7=9A=84=E7=AE=AD=E5=BC=95=E7=88=86=E6=B0=94?=
=?UTF-8?q?=E7=90=83.md?=
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Added python version code
---
.../0452.用最少数量的箭引爆气球.md | 18 +++++++++++++++---
1 file changed, 15 insertions(+), 3 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 f62fb153..45c94a01 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
@@ -70,7 +70,7 @@
其实都可以！只不过对应的遍历顺序不同，我就按照气球的起始位置排序了。
-既然按照其实位置排序，那么就从前向后遍历气球数组，靠左尽可能让气球重复。
+既然按照起始位置排序，那么就从前向后遍历气球数组，靠左尽可能让气球重复。
从前向后遍历遇到重叠的气球了怎么办？
@@ -167,7 +167,19 @@ class Solution {
```
Python：
-
+```python
+class Solution:
+ def findMinArrowShots(self, points: List[List[int]]) -> int:
+ if len(points) == 0: return 0
+ points.sort(key=lambda x: x[0])
+ result = 1
+ for i in range(1, len(points)):
+ if points[i][0] > points[i - 1][1]: # 气球i和气球i-1不挨着，注意这里不是>=
+ result += 1
+ else:
+ points[i][1] = min(points[i - 1][1], points[i][1]) # 更新重叠气球最小右边界
+ return result
+```
Go：
@@ -178,4 +190,4 @@ Go：
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频：[代码随想录](https://space.bilibili.com/525438321)
* 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
\ No newline at end of file
+
