增加0452.用最少数量的箭引爆气球，python，基于右侧坐标排序的算法，节约表达式中的判断步骤

problems/0452.用最少数量的箭引爆气球.md

@@ -200,6 +200,29 @@ class Solution: # 不改变原数组
                 curr_min_right = min(curr_min_right, i[1])
         return count
 ```
+```python
+class Solution: # 不改变原数组，基于右侧坐标排序
+    def findMinArrowShots(self, points: List[List[int]]) -> int:
+        # 题目说明points列表长度至少为1
+        if len(points) == 1:
+            return 1
+
+        # 根据右侧坐标排序
+        points.sort(key=lambda x: x[1])
+
+        # 初始化
+        arrow = 0
+        curr_min_right = float('-inf')
+
+        # 根据右侧坐标排序后，更新curr_min_right时的point[1]自动保证为剩余气球中最小的
+        # 右侧坐标值，节约else表达式中的判断步骤
+        for point in points:
+            if point[0] > curr_min_right:
+                arrow += 1
+                curr_min_right = point[1]
+
+        return arrow
+```
 ### Go
 ```go
 func findMinArrowShots(points [][]int) int {