[Leetcode 42, Hard] Trapping Rain Water

Problem:

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!

Analysis:

Solutions:

C++:

    int trap(vector<int>& height) {        if(height.empty())            return 0;                    stack<int> end_point_stack;        for(int i = 0; i < height.size(); ++i) {            if(height[i] == 0)                continue;            else {                if((i == 0 && height[i] > height[i + 1]) || (i == height.size() - 1 && height[i] > height[i - 1])                   || (height[i] >= height[i + 1] && height[i] > height[i - 1])                    || (height[i] > height[i + 1] && height[i] >= height[i - 1]))                     {                        if(end_point_stack.size() <= 1 || height[end_point_stack.top()] >= height[i])                            end_point_stack.push(i);                        else {                            while(end_point_stack.size() > 1) {                                int top_height = end_point_stack.top();                                end_point_stack.pop();                                if(height[top_height] > height[end_point_stack.top()] || height[top_height] >= height[i]) {                                    end_point_stack.push(top_height);                                    end_point_stack.push(i);                                    break;                                }                            }                                                        if(end_point_stack.size() == 1)                                end_point_stack.push(i);                        }                    }            }        }                if(end_point_stack.empty())            return 0;                vector<int> index;        while(!end_point_stack.empty()) {            index.insert(index.begin(), end_point_stack.top());            end_point_stack.pop();        }                int r_sum = 0;        for(int i = 0; i < index.size() - 1; ++i) {            int h = min(height[index[i]], height[index[i + 1]]);            int part_sum = 0;            for(int j = index[i] + 1; j <  index[i + 1]; ++j) {                if(height[j] > h)                    continue;                part_sum += h - height[j];            }            r_sum += part_sum;        }                return r_sum;    }

Java:

Python: