目录

• 题目截图
• 题目分析
• ac code
• 总结

题目截图

题目分析

• 找到k的位置
• 然后一步步往左走，一步步往右走
• 统计左边和右边的比当前k小的和比k大的
• lst = [[small, big]]，分为left和right两部分
• 可以先一侧的单独看small和big，找到big - small = 0或者1的即可
• 如果两侧的话，比较麻烦
• 我们将left和right两侧的再转化成diff = big - small
• 那么我们需要在左侧找一个diff1和右侧找一个diff2
• 使得diff1 + diff2 = 0或者1即可
• 特别地，我们固定diff1，然后可以得到diff2两个可能的值
• 因此，可以先对diff2排序，再用bisect_left和bisect_right卡住所有可选的值即可

ac code

class Solution:def countSubarrays(self, nums: List[int], k: int) -> int:n = len(nums)k_idx = 0p = [0] * (n + 1)for i, v in enumerate(nums):p[v] = iwhile nums[k_idx] != k:k_idx += 1# s个比k小，s个比k大# s个比k小，s + 1个比k大left, right = k - 1, n - klefts, rights = [], []# 左边的情况small, big = 0, 0for i in range(k_idx - 1, -1, -1):if nums[i] < k:small += 1else:big += 1lefts.append([small, big])# 右边的情况small, big = 0, 0for i in range(k_idx + 1, n):if nums[i] < k:small += 1else:big += 1rights.append([small, big])ans = 1 # 自己# small - big#print(lefts)#print(rights)lefts_diff = []for small, big in lefts:lefts_diff.append(big - small)rights_diff = []for small, big in rights:rights_diff.append(big - small)# print(lefts_diff)# print(rights_diff)# 两个和要等于0或者1ans = 1 # 自己for diff in lefts_diff:if diff == 0 or diff == 1:ans += 1for diff in rights_diff:if diff == 0 or diff == 1:ans += 1rights_diff.sort()for num in lefts_diff:need1 = -numneed2 = 1 - numi = bisect_left(rights_diff, need1)j = bisect_right(rights_diff, need2)ans += j - ireturn ans

总结

• 一个关键点：比k大的 - 比k小的 = 0或者1
• 为了研究子数组，我们可以考虑记录左侧连续的和右侧连续的出现的small和big个数
• 我们需要big1 + big2 - small1 - small2 = 0或者1
• 直接先变成diff1和diff2使得，diff1 + diff2 = 0或者1
• 那就好办了
• 遍历diff1，排序diff2，用二分卡住两边，得到数量即可