動機
atMost或是在回收時計算長度
Problem
Given an array of integers nums
and an integer k
. A continuous subarray is called nice if there are k
odd numbers on it.
Return the number of nice sub-arrays.
Example 1:
Input: nums = [1,1,2,1,1], k = 3Output: 2Explanation: The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].
Example 2:
Input: nums = [2,4,6], k = 1Output: 0Explanation: There is no odd numbers in the array.
Example 3:
Input: nums = [2,2,2,1,2,2,1,2,2,2], k = 2Output: 16
Constraints:
1 <= nums.length <= 50000
1 <= nums[i] <= 10^5
1 <= k <= nums.length
Sol
class Solution:
def numberOfSubarrays(self, nums: List[int], k: int) -> int:
def atmost(limit):
i = cnt = ret = 0
for j,n in enumerate(nums):
if n % 2 == 1:
cnt += 1
while i <= j and cnt > limit:
if nums[i] % 2 == 1:
cnt -= 1
i += 1
ret += j-i+1
return ret
return atmost(k)-atmost(k-1)
另一個是在回收時計算長度
def numberOfSubarrays(self, A, k):
i = count = res = 0
for j in range(len(A)):
if A[j] & 1:
k -= 1
count = 0 # 等累積到合法時才計算長度
while k == 0:
k += A[i] & 1
i += 1
count += 1
res += count
return res