【Leetcode】Trapping Rain Water

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. 

class Solution {
public:
    int trap(int A[], int n) {
        int sum = 0;
        int *max_left = new int[n]();
        for (int i = 1; i < n - 1; ++i) {
            if (A[i - 1] > max_left[i - 1]) {
                max_left[i] = A[i - 1];
            } else {
                max_left[i] = max_left[i - 1];
            }
        }
        int max_right = 0;
        for (int i = n - 2; i >= 1; --i) {
            if (A[i + 1] > max_right) {
                max_right = A[i + 1];
            }
            if (A[i] < max_left[i] && A[i] < max_right) {
                sum += (min(max_left[i], max_right) - A[i]);
            }
        }
        return sum;
    }
};

单独考虑每根bar，每根bar上方可以盛的水量取决于从它向左扫描到的最高的bar和向右扫描到的最高的bar中较矮的那一个和它本身的高度差。

扫描一遍数组，记录下每根bar左边最高的bar，然后从右向左，一遍记录当前最高的bar，一边计算每根bar上方可以盛水的量，并求和。