[LeetCode] Trapping Rain Water

Trapping Rain Water

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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!

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我的分析是這樣的，每一列中可以存儲的水量爲它的左右兩邊的最高的高度中較小的一個與該列的高度的差值，如果差值大於0就可以存儲水，否則就不可以存儲水，最後

所有的水量之和想家就可以了。

好了直接上代碼：

class TrappRainWater
{
public:
    int trap(int A[], int n) {
        if(n==0) return 0;
        int *leftMaxArray = new int[n];
        int *rightMaxArray = new int[n];
        leftMaxArray[0] = 0;
        rightMaxArray[n-1] = 0;
        int max = A[0];
        for(int i=1;i<n;i++)
        {
            leftMaxArray[i] = max;
            max = max<A[i]?A[i]:max;
        }
        max = A[n-1];
        for(int i=n-2;i>=0;i--)
        {
            rightMaxArray[i] = max;
            max = max<A[i]?A[i]:max;
        }
        int ret = 0;
        for(int i=0;i<n;i++)
        {
            int smallValue = leftMaxArray[i]<rightMaxArray[i]?leftMaxArray[i]:rightMaxArray[i];
            if(smallValue>A[i])
                ret+=(smallValue-A[i]);
        }
        return ret;
    }
};