题目：
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!
题解：
参考：低调小一（http://blog.csdn.net/wzy_1988/article/details/17752809）的解题思路
water的容量，取决于A[i]左右两边的高度（可延展）较小值与A[i]的差值，即volume[i] = [min(left[i],
right[i]) - A[i]] * 1，这里的1是宽度，如果the width of each bar is 2,那就要乘以2了”
那么如何求A[i]的左右高度呢？ 要知道，能盛多少水主要看短板。那么对每个A[i]来说，要求一个最高的左短板，再求一个最高的右短板，这两个直接最短的板子减去A[i]原有的值就是能成多少水了。
所以需要两遍遍历，一个从左到右，找最高的左短板；一个从右到左，找最高的右短板。最后记录下盛水量的总值就是最终结果了。
代码如下：
1 public int trap(int[] A) {
2 if (A == null || A.length == 0)
3 return 0;
4
5 int i, max, total = 0;
6 int left[] = new int[A.length];
7 int right[] = new int[A.length];
8
9 // from left to right
left[0] = A[0];
max = A[0];
for (i = 1; i < A.length; i++) {
left[i] = Math.max(max, A[i]);
max = Math.max(max, A[i]);
}
// from right to left
right[A.length-1] = A[A.length-1];
max = A[A.length-1];
for (i = A.length-2; i >= 0; i--) {
right[i] = Math.max(max, A[i]);
max = Math.max(max, A[i]);
}
// trapped water (when i==0, it cannot trapped any water)
for (i = 1; i < A.length-1; i++) {
int bit = Math.min(left[i], right[i]) - A[i];
if (bit > 0)
total += bit;
}
return total;
}
对照着代码再看原来的例子：
index: 0 1 2 3 4 5 6 7 8 9 10 11
A[index]: 0 1 0 2 1 0 1 3 2 1 2 1
left[index]: 0 1 1 2 2 2 2 3 3 3 3 3
right[index]: 3 3 3 3 3 3 3 3 2 2 2 1
min[i]: 0 1 1 2 2 2 2 3 2 2 2 1
bit[i]: - 0 1 0 1 2 1 0 0 1 0 0
那么根据上表可以算出最终结果是6。
Reference：http://blog.csdn.net/wzy_1988/article/details/17752809