【题解】42. 接雨水（动态规划 预处理）

https://leetcode.cn/problems/trapping-rain-water/description/

class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();

        // 预处理数组
        vector<int> lefts(n, 0);
        vector<int> rights(n, 0);

        // 预处理记录左侧最大值
        lefts[0] = height[0];
        for (int i = 1; i < n; ++i) {
            lefts[i] = max(lefts[i-1], height[i]);
        }

        // 预处理记录右侧最大值
        rights[n-1] = height[n-1];
        for (int i = n - 2; i > 0; --i) {
            rights[i] = max(rights[i+1], height[i]);
        }

        // 将每个柱子可以接到水的值进行累加
        int ret = 0;
        for (int i = 0; i < n; ++i) 
        {
            if (min(lefts[i], rights[i]) - height[i] > 0)
                ret += min(lefts[i], rights[i]) - height[i];
        }    

        return ret;
    }
};