# 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!

Solution I:

``````public class Solution {
public int trap(int[] height) {
if (height == null || height.length == 0) {
return 0;
}

int[] left = new int[height.length];
left[0] = height[0];

for (int i = 1; i < height.length; i++) {
left[i] = Math.max(left[i-1], height[i]);
}

int[] right = new int[height.length];
right[height.length-1] = height[height.length-1];

for (int i = height.length-2; i >= 0; i--) {
right[i] = Math.max(right[i+1], height[i]);
}

int trappedWater = 0;

for (int i = 0; i < height.length; i++) {
int curr = Math.min(left[i], right[i]);
if (curr > height[i]) {
trappedWater +=  (curr - height[i]);
}
}

return trappedWater;
}
}
``````

Solution II:

http://blog.csdn.net/linhuanmars/article/details/20888505

``````public int trap(int[] A) {
if(A==null || A.length ==0)
return 0;
int l = 0;
int r = A.length-1;
int res = 0;
while(l<r)
{
int min = Math.min(A[l],A[r]);
if(A[l] == min)
{
l++;
while(l<r && A[l]<min)
{
res += min-A[l];
l++;
}
}
else
{
r--;
while(l<r && A[r]<min)
{
res += min-A[r];
r--;
}
}
}
return res;
}
``````