42. 接雨水

2021/03/23 leetcode 共 1007 字,约 3 分钟

title

problem link

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:

img

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.length
  • 0 <= n <= 3 * 104
  • 0 <= 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

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