[leet code] 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.

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Analysis:

Key point of this problem is to find out the logic to compute water trapped in each "container".  

In order to trap the water, there has to be a left bound, a right bound.  My original idea was to calculate the amount of trapped water element by element from left bound of the container till the right bound reached.  For example, trapped water between Array[3] -> Array[7] (example of the problem) is (2-1) + (2-0) + (2-1) = 4.  

From this example, we can also find that left bound of this "container" is the previous highest value (Array[3]=2); right bound of the container is the current highest value (Array[7]=3).  In the next step, we can move the left bound to Array[7], and repeat this computation till the next right bound (i.e. next highest bound) reached.

However, this approach ignore the water trapped in Array[9].  Because when the left bound reached the highest value (assume it is the only highest value) in the array, there would not be right bound anymore to construct a "container".  Therefore, the approach above can only compute the water trapped from left most to the highest value.  

To complet the whole computation, we can simply repeat the approach one  more time from right most to the highest value.  

public class Solution {    public int trap(int[] A) {        if (A.length <= 2) return 0;                int singleSum = 0; // one hole        int totalSum = 0; // result                        // from left to right        int left = 0;        for(int i=1; i<A.length; i++){            if(A[left]>A[i]) singleSum +=A[left]-A[i];            else {                totalSum += singleSum;                left = i;                singleSum = 0;            }        }                // from right to left        singleSum = 0; // reset singleSum        int right = A.length-1;        for (int i=A.length-2; i>=left; i--){            if(A[right]>A[i]) singleSum += A[right]-A[i];            else{                totalSum += singleSum;                right = i;                singleSum = 0;            }        }                return totalSum;    }}