title
给定 n 个非负整数表示每个宽度为 1 的柱子的高度图,计算按此排列的柱子,下雨之后能接多少雨水。
示例 1:

输入:height = [0,1,0,2,1,0,1,3,2,1,2,1]
输出:6
解释:上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图,在这种情况下,可以接 6 个单位的雨水(蓝色部分表示雨水)。
示例 2:
输入:height = [4,2,0,3,2,5]
输出:9
提示:
n == height.length0 <= n <= 3 * 1040 <= height[i] <= 105
solution
单调栈, 参考
class Solution:
def trap(self, height: List[int]) -> int:
if not height:
return 0
stack = []
ans = 0
for i, h in enumerate(height):
while stack and h > height[stack[-1]]:
popindex = stack.pop()
while stack and height[popindex] == height[stack[-1]]:
stack.pop()
if stack:
tmp = height[stack[-1]] # stack value
hv = min(tmp - height[popindex], h - height[popindex])
width = i - stack[-1] - 1
ans += hv * width
stack.append(i)
return ans
常规解法
class Solution:
def trap(self, height: List[int]) -> int:
n = len(height)
left = [0 for _ in range(n)]
right = [0 for _ in range(n)]
max_left = 0
for i, v in enumerate(height):
left[i] = max_left
max_left = max(max_left, v)
max_right = 0
for i in range(n - 1, -1, -1):
v = height[i]
right[i] = max_right
max_right = max(max_right, v)
res = 0
for i, v in enumerate(height):
t = min(left[i], right[i])
if t > v:
res += t - v
return res