LeetCode：Trapping Rain Water

算法很简单，核心思想是：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，在第二遍的同时就可以计算出i位置的结果了，而且并不需要开空间来存放rightMostHeight数组。

时间复杂度是O(n)，只扫了两遍。

class Solution {public:    int trap(int A[], int n) {        //no way to contain any water        if(n <= 2) return 0;                //1. first run to calculate the heiest bar on the left of each bar        int *lmh = new int[n];//for the most height on the left        lmh[0] = 0;        int maxh = A[0];        for(int i = 1; i < n; ++i) {            lmh[i] = maxh;            if(maxh < A[i]) maxh = A[i];        }        int trapped = 0;                //2. second run from right to left,         // caculate the highest bar on the right of each bar        // and calculate the trapped water simutaniously        maxh = A[n-1];        for(int i = n - 2; i > 0; --i) {            int left = lmh[i];            int right = maxh;            int container = min(left,right);            if(container > A[i]) {                trapped += container - A[i];            }            if(maxh < A[i]) maxh = A[i];        }        delete lmh;        return trapped;    }};