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🐱(dynamic): 898. 子数组按位或操作

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`@@ -725,4 +725,26 @@ class Solution(object):`
`725`	`725`	`# 选择买入或不变`
`726`	`726`	`hold = max(hold, cash - prices[i])`
`727`	`727`	`return cash`
	`728`	+```
	`729`	`+`
	`730`	`+## 898. 子数组按位或操作`
	`731`	`+`
	`732`	`+[原题链接](https://leetcode-cn.com/problems/bitwise-ors-of-subarrays/)`
	`733`	`+`
	`734`	`+### 思路`
	`735`	`+`
	`736`	+动态规划,`dp[i]` 存储所有以 `i` 结尾的子数组的或结果(集合)。
	`737`	`+`
	`738`	+由于数据规模为 `1 <= A.length <= 50000`,因此无法使用 `O(n^2)` 的算法。参考 [花花酱 LeetCode 898. Bitwise ORs of Subarrays - 刷题找工作 EP222](https://www.bilibili.com/video/av31142168) 的讲解,说的非常详细了,也大致了解到如何根据问题规模确定复杂度了。
	`739`	`+`
	`740`	+```python
	`741`	`+class Solution:`
	`742`	`+ def subarrayBitwiseORs(self, A: List[int]) -> int:`
	`743`	`+ cur = set()`
	`744`	`+ res = set()`
	`745`	`+ # cur 存储以 a 结尾的所有子数组的或结果`
	`746`	`+ for a in A:`
	`747`	`+ cur = {n \| a for n in cur} \| {a}`
	`748`	`+ res \|= cur`
	`749`	`+ return len(res)`
`728`	`750`	```

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