42. 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.
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!
Example:
Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Solution
(1) Java
class Solution {
public int trap(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
int[] left = new int[height.length];
int[] right = new int[height.length];
left[0] = height[0];
right[height.length-1] = height[height.length-1];
for (int i = 1; i < height.length; i++) {
left[i] = Math.max(left[i-1], height[i]);
}
for (int j = height.length-2; j >= 0; j--) {
right[j] = Math.max(right[j+1], height[j]);
}
int rst = 0;
for (int i = 0; i < height.length; i++) {
rst += Math.min(left[i],right[i]) - height[i];
}
return rst;
}
}
(2) Python
class Solution:
def trap(self, height):
"""
:type height: List[int]
:rtype: int
"""
if not height:
return 0
rst = 0
left, right = 0, len(height)-1
while left < right:
if height[left] <= height[right]:
move = left+1
while move < right and height[move] <= height[left]:
rst += height[left]-height[move]
move += 1
left = move
else:
move = right-1
while move > left and height[move] <= height[right]:
rst += height[right]-height[move]
move -= 1
right = move
return rst
(3) Scala