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.
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];
int max = height[0];
for (int i = 1; i < height.length; i++) {
max = Math.max(max, height[i]);
left[i] = max;
}
int[] right = new int[height.length];
right[height.length - 1] = height[height.length - 1];
max = height[height.length - 1];
for (int i = height.length - 2; i >= 0; i--) {
max = Math.max(max, height[i]);
right[i] = max;
}
int sum = 0;
for (int i = 0; i < height.length; i++) {
int min = Math.min(left[i], right[i]);
if (min - height[i] > 0) {
sum += min - height[i];
}
}
return sum;
}
}
Hope this helps,
Michael