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.

Trapping Rain Water
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.

Two Pointer

这种做法很难想到，感觉有点贪心的味道。

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public int trap(int[] heights) {
    if (heights == null || heights.length == 0)
        return 0;
    int left = 0, right = heights.length - 1;
    int ans = 0;
    int maxLeft = 0, maxRight = 0;
    
    while (left < right) {
        if (heights[left] <= heights[right]) {
            if (heights[left] > maxLeft)
                maxLeft = heights[left];
            else
                ans += maxLeft - heights[left];
            left++;
        } else {
            if (heights[right] > maxRight)
                maxRight = heights[right];
            else
                ans += maxRight - heights[right];
            right--;
        }
    }
    return ans;
}