[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. Thanks Marcos for contributing this image!

仔细想想感觉这道题应该是扫一遍就能得到结果的。。。对某个值A[i]来说，能trapped的最多的water取决于在i之前最高的值leftMostHeight[i]和在i右边的最高的值rightMostHeight[i]（均不包含自身）。如果min(left,right) > A[i]，那么在i这个位置上能trapped的water就是min(left,right) – A[i]。有了这个想法就好办了，第一遍从左到右计算数组leftMostHeight，第二遍从右到左计算rightMostHeight。时间复杂度是O(n)。

class Solution {
public:
    int trap(int A[], int n) {
        int maxHeight = 0;
        vector<int> leftMostHeight(n, 0);
        for (int i = 0; i < n; ++i) {
            leftMostHeight[i] = maxHeight;
            maxHeight = (maxHeight > A[i]) ? maxHeight : A[i];
        }
        maxHeight = 0;
        vector<int> rightMostHeight(n, 0);
        for (int i = n - 1; i >= 0; --i) {
            rightMostHeight[i] = maxHeight;
            maxHeight = (maxHeight > A[i]) ? maxHeight : A[i];
        }
        int water = 0, height;
        for (int i = 0; i < n; ++i) {
            height = min(leftMostHeight[i], rightMostHeight[i]) - A[i];
            water += (height > 0) ? height: 0;
        }
        return water;
    }
};

 下面是扫一遍的代码：

class Solution {
public:
    int trap(int A[], int n) {
        int left = 0, right = n - 1;
        int height = 0, res = 0;
        while (left < right) {
            if (A[left] < A[right]) {
                height = max(height, A[left]);
                res += height - A[left];
                ++left;
            } else {
                height = max(height, A[right]);
                res += height - A[right];
                --right;
            }
        }
        return res;
    }
};