1. 题目

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!

2. 思路

每次以第一个点为起点，找到后续第一个比起点大于等于的点作为终点。下一次的起点就是上一次的终点。
如果找终点时直到找到末尾也没找到，就将终点设置为当前查找段的最大的点。

确定好段之后，段的首尾是段内的最大和次大点，则直接计算首尾的最大容量，再减去内部的填充点占用即可。

3. 代码

耗时：12ms

class Solution {
public:
    // 划分为一段段的处理，每一段是从起点开始，终点是第一个大于等于起点的点。
    // 如果终点小于起点，则回退到段内的非起点最高点，作为一段。
    int trap(vector<int>& height) {
        if (height.size() < 3) {return 0; }
        int sum = 0;
        int start = 0;
        int ls = height[start];
        int end = start + 1;
        int max_end = start + 1; // start之后, end之前的最大点下标
        int le = 0;
        while (end < height.size()) {
            int cle = height[end];
            if (cle >= ls) {
                sum += trap(height, start, end);
                start = end;
                ls = cle;
                end = start + 1;
                le = 0;
                max_end = end;
                continue;
            } else if (cle > le) {
                le = cle;
                max_end = end;
            }
            ++end;
            if (end == height.size()) {
                end = max_end;
                sum += trap(height, start, end);
                start = end;
                end = start + 1;
                le = 0;
                max_end = end;
                
            }
        }
        
        return sum;
    }
    
    int trap(vector<int>& height, int start, int end) {
        //cout << "s=" << start << " e=" << end << endl;
        if (end - start < 2) {return 0;}
        int sum = (end - start - 1) * min(height[start], height[end]);
        for (int i = start + 1; i < end; i++) {
            sum -= height[i];
        }
        return sum;
    }
};