leetcode(30)单调栈
6077. 巫师的总力量和
注意：因为要求连续，所以不能用回溯的方法做
496. 下一个更大元素 I
907. 子数组的最小值之和
class Solution:
def sumSubarrayMins(self, arr: List[int]) -> int:
res = 0
stack = []
arr = [-inf] + arr + [-inf]
for r, a in enumerate(arr):
while stack and arr[stack[-1]] > a:
i = stack.pop()
res += arr[i] * (i - stack[-1]) * (r - i)
stack.append(r)
return res % (10**9 + 7)
1856. 子数组最小乘积的最大值
前缀和+单调栈
class Solution:
def maxSumMinProduct(self, nums: List[int]) -> int:
pre_sum = [0]
for i in range(len(nums)):
pre_sum.append(nums[i] + pre_sum[i])
res = 0
nums = [-inf] + nums + [-inf]
stack = []
for r, n in enumerate(nums):
while stack and nums[stack[-1]] > n:
i = stack.pop()
# print(nums[i])
# print(nums[stack[-1]:r])
cur = nums[i] * (pre_sum[r - 1] - pre_sum[stack[-1]]) # 注意r-1
res = max(res, cur)
stack.append(r)
return res % (10 ** 9 + 7)
2104. 子数组范围和
使用两次单调栈，最大值和最小值分开求
class Solution:
def subArrayRanges(self, nums: List[int]) -> int:
res = 0
new_nums = [-inf] + nums + [-inf]
stack = []
for r, n in enumerate(new_nums):
while stack and new_nums[stack[-1]] > n:
i = stack.pop()
res -= new_nums[i] * (r - i) * (i - stack[-1])
stack.append(r)
new_nums = [inf] + nums + [inf]
stack = []
for r, n in enumerate(new_nums):
while stack and new_nums[stack[-1]] < n:
i = stack.pop()
res += new_nums[i] * (r - i) * (i - stack[-1])
stack.append(r)
return res
901. 股票价格跨度
用单调栈维护一个单调递减的价格序列，并且对于每个价格，存储一个 weight 表示它离上一个价格之间（即最近的一个大于它的价格之间）的天数。
如果是栈底的价格，则存储它本身对应的天数。例如 [11, 3, 9, 5, 6, 4, 7] 对应的单调栈为 (11, weight=1), (9, weight=2), (7, weight=4)。
class StockSpanner:
def __init__(self):
self.stack = []
def next(self, price: int) -> int:
weight = 1
while self.stack and self.stack[-1][0] <= price:
weight += self.stack.pop()[1]
self.stack.append((price, weight))
return weight
173. 二叉搜索树迭代器
先把根节点及其左子树存入，然后依次出栈；每次调用next判断有右子树，则同样入栈
class BSTIterator:
def __init__(self, root: Optional[TreeNode]):
self.stack = []
while root:
self.stack.append(root)
root = root.left
def next(self) -> int:
cur = self.stack.pop()
node = cur.right
while node:
self.stack.append(node)
node = node.left
return cur.val
def hasNext(self) -> bool:
return len(self.stack) > 0
