2.1.14 TrappingRainWater

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Notes:
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.

Solution: 1. Find left bound and right bound for each element. O(n).
2. more space efficiency. Time: O(n), Space: O(1);对于每个柱子，找到其左右两边最高的柱子，该柱子能容纳的面积就是min(max_left,
max_right) - height。在求柱子的左右最大值的时候，包括柱子本身。

class Solution {public:    int trap_1(int A[], int n) {        if (n == 0) return 0;        vector<int> maxLeft(n,0);        vector<int> maxRight(n,0);        maxLeft[0] = A[0];        maxRight[n - 1] = A[n - 1];        for (int i = 1; i < n; ++i) {            maxLeft[i] = max(maxLeft[i - 1], A[i]);            maxRight[n - 1 - i] = max(maxRight[n - i], A[n - 1 - i]);        }                int res = 0;        for (int i = 1; i < n; ++i) {            res += min(maxLeft[i], maxRight[i]) - A[i];        }        return res;    }}

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