【34.7】【H】【leetcode】Trapping Rain Water

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example,

Given

[0,1,0,2,1,0,1,3,2,1,2,1]

, return

6

.



The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

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对某个值A[i]来说，能trapped的最多的water取决于在i之前最高的值leftMostHeight[i]和在i右边的最高的值rightMostHeight[i]（均不包含自身）。

如果min(left,right) > A[i]，那么在i这个位置上能trapped的water就是min(left,right) – A[i]。

有了这个想法就好办了，第一遍从左到右计算数组leftMostHeight，第二遍从右到左计算rightMostHeight。

时间复杂度是O(n)。

class Solution(object):
def trap(self, height):

l = len(height)
h = height
if l < 2:
return 0

maxx = h[0]

left_max = [0]*l
right_max = [0]*l

left_max[0] = -1#h[0]
for i in range(1,l):
left_max[i] = maxx
maxx = max(maxx,h[i])

maxx = h[l-1]
right_max[l-1] = -1#h[l-1]
for i in range(l-2,-1,-1):
right_max[i] = maxx
maxx = max(maxx,h[i])

#print left_max
#print right_max

res = 0
for i in range(0,l):
t = min(left_max[i],right_max[i]) - h[i]
if t > 0:
res += t
return res