Commit a7f3cb3

committed

🐱(sliding-window): 1248. 统计「优美子数组」

1 parent aaa8cdc commit a7f3cb3Copy full SHA for a7f3cb3

File tree

+31

-1

lines changed

+31

-1

lines changed

Lines changed: 31 additions & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -212,4 +212,34 @@ class Solution:`
`212`	`212`	```
`213`	`213`
`214`	`214`	`- 时间复杂度:$O(l1 + (l2 - l1))$($l1$ 为字符串 s1 长度,$l2$ 为字符串 s2 长度)`
`215`		`-- 空间复杂度:$O(1)$`
	`215`	`+- 空间复杂度:$O(1)$`
	`216`	`+`
	`217`	`+## 1248. 统计「优美子数组」`
	`218`	`+`
	`219`	`+[原题链接](https://leetcode-cn.com/problems/count-number-of-nice-subarrays/)`
	`220`	`+`
	`221`	`+### 滑动窗口`
	`222`	`+`
	`223`	`+记录奇数的位置。固定 k 个奇数,子数组的个数 = 第一个奇数左边偶数的个数 * 最后一个奇数右边偶数的个数。`
	`224`	`+`
	`225`	+```python
	`226`	`+class Solution:`
	`227`	`+ def numberOfSubarrays(self, nums: List[int], k: int) -> int:`
	`228`	`+ ans = 0`
	`229`	`+ # 开始位置`
	`230`	`+ odd = [-1]`
	`231`	`+ # 记录奇数下标`
	`232`	`+ for i in range(len(nums)):`
	`233`	`+ if nums[i] % 2 == 1:`
	`234`	`+ odd.append(i)`
	`235`	`+ # 添加结束位置`
	`236`	`+ odd.append(len(nums))`
	`237`	`+`
	`238`	`+ # 遍历奇数数组`
	`239`	`+ for j in range(1, len(odd) - k):`
	`240`	`+ # 从第 i 个到 i + k - 1`
	`241`	`+ # 第 i 个奇数前面的偶数个数 * 第 i + k - 1 后面的偶数个数`
	`242`	`+ ans += (odd[j] - odd[j - 1]) * (odd[j + k] - odd[j + k - 1])`
	`243`	`+`
	`244`	`+ return ans`
	`245`	+```

Comments

(0)