题目：https://leetcode-cn.com/problems/trapping-rain-water/
给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。

我的方法一：双指针

步骤

虽然官方解法也是双指针，但是明显官方的解法的更简洁，所以主要看看官方的方法，https://leetcode-cn.com/problems/trapping-rain-water/solution/jie-yu-shui-by-leetcode-solution-tuvc/

class Solution {
public:
    int trap(vector<int>& height) {
        int ans = 0;
        int left = 0, right = height.size() - 1;
        int leftMax = 0, rightMax = 0;
        while (left < right) {
            leftMax = max(leftMax, height[left]);
            rightMax = max(rightMax, height[right]);
            if (height[left] < height[right]) {
                ans += leftMax - height[left];
                ++left;
            } else {
                ans += rightMax - height[right];
                --right;
            }
        }
        return ans;
    }
};

我的解法麻烦

class Solution {
public:
    int trap(vector<int>& height) {
        if(height.size()<3){
            return 0;
        }

        int sum = 0;
        int left = 0;
        int right = height.size()-1;
        int left_height=0;
        int right_height=0;
        int lower_height;
        int scanned_height=0;
        while(left<right){
            left_height = height[left];
            right_height = height[right];
            lower_height = left_height<right_height?left_height:right_height;

            if(lower_height<=scanned_height){
                if(left_height>right_height){
                    right--;
                }else{
                    left++;
                }
                continue;
            }

            for(int i = left+1; i<right; i++){
                if(height[i]<lower_height){
                    if(height[i]>=scanned_height){
                        sum += lower_height-height[i];
                    }else{
                        sum += lower_height-scanned_height;
                    }
                }
            }

            if(left_height>right_height){
                right--;
            }else{
                left++;
            }
            scanned_height=lower_height;
        }

        return sum;
    }
};

我的方法二：动态规划