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🐱(greedy): 452. 用最少数量的箭引爆气球

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`@@ -268,6 +268,39 @@ class Solution:`
`268`	`268`	`return people`
`269`	`269`	```
`270`	`270`
	`271`	`+## 452. 用最少数量的箭引爆气球`
	`272`	`+`
	`273`	`+[原题链接](https://leetcode-cn.com/problems/minimum-number-of-arrows-to-burst-balloons/)`
	`274`	`+`
	`275`	`+### 思路`
	`276`	`+`
	`277`	+1. 对 `points` 按 `begin` 进行升序排序
	`278`	+2. 循环遍历 `points` 判断相邻位置的区间是否发生重叠
	`279`	+ - 如果重叠:说明只需要一支箭就可以射穿,并把重叠区间的 `end` 值记录下来,用于比较下一次的区间重叠情况
	`280`	`+ - 如果不重叠:需要射出的箭数量 + 1`
	`281`	`+`
	`282`	+```python
	`283`	`+class Solution:`
	`284`	`+ def findMinArrowShots(self, points: List[List[int]]) -> int:`
	`285`	`+ length = len(points)`
	`286`	`+ if length == 0:`
	`287`	`+ return 0`
	`288`	`+ res = 1`
	`289`	`+ # 排序`
	`290`	`+ points.sort(key=lambda x: x[0])`
	`291`	`+`
	`292`	`+ end = points[0][1]`
	`293`	`+ # 遍历`
	`294`	`+ for i in range(1, length):`
	`295`	`+ if points[i][0] > end:`
	`296`	`+ res += 1`
	`297`	`+ end = points[i][1]`
	`298`	`+ else:`
	`299`	`+ end = min(end, points[i][1])`
	`300`	`+`
	`301`	`+ return res`
	`302`	+```
	`303`	`+`
`271`	`304`	`## 621. 任务调度器`
`272`	`305`
`273`	`306`	`[原题链接](https://leetcode-cn.com/problems/task-scheduler/)`

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