2.1.15 Trapping Rain Water

Link: https://oj.leetcode.com/problems/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!
我的思路：remember that 对每一个A[i], 它存储的水量等于它的左右两边的最大值， max(left), max(right)，取两者的最小值，然后减去A[i].
min(max(left), max(right)) - A[i]

问题是max(left)可以通过从左向右循环得到，max(right）怎么得到？

Approach I: 三次遍历。1 从左往右，对每个柱子，找左边最大值。2 从右往左，对每个柱子，找右边最大值。3 求每个柱子上面的面积并累加。

我的代码：一次过。可以不再做。

Time: O(n), Space: O(1)

public class Solution {
public int trap(int[] A) {
if(A.length <=2) return 0;
//get max left
int[] maxL = new int[A.length];
maxL[0] = 0;
for(int i = 1; i < A.length; i++){
maxL[i] = Math.max(maxL[i-1], A[i-1]);
}
//get max right
int[] maxR = new int[A.length];
maxR[A.length-1] = 0;
for(int i = A.length-2; i >=0; i--){
maxR[i] = Math.max(maxR[i+1], A[i+1]);
}
//get water
int sum = 0;
for(int i = 0; i < A.length; i++){
int diff = Math.min(maxL[i], maxR[i])-A[i];
if(diff >0){
sum += diff;
}
}
return sum;
}
}
Note: Don't forget this! Otherwise has Runtime Error. 

if(A.length <=2) return 0;

Approach II: 把第二，三次遍历合起来。
public class Solution {
public int trap(int[] A) {
int n = A.length;
if(n <=2) return 0;
//get max left
int[] maxL = new int
;
maxL[0] = 0;
for(int i = 1; i < n; i++){
maxL[i] = Math.max(maxL[i-1], A[i-1]);
}
//combine get max right and sum together
int sum = 0;
int maxR = A[n-1];
for(int i = n-2; i >=0; i--){
int diff = Math.min(maxL[i], maxR)-A[i];
if(diff > 0){
sum += diff;
}
if(A[i] > maxR){
maxR = A[i];
}
}
return sum;
}
}