戳我去解题

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. 

解题分析：

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

我们从前往后遍历，对于每个柱子，我们可以时刻更新 其左边的最大值，但是 其右边的最大值，我们可以一次次的进行 std::max_element

但是这样  T(n) = O(n^2)

我们可以用空间换时间，首先用一个数组保存 每个柱子右边的最大值，这样 T(n) = O(n)

1. class Solution {
   public:
       int trap(int A[], int n) {
           assert(A != NULL && n >= 0);
           if (n <= 2) return 0;
           int leftMax = A[0];
           int rightMax[n];
           rightMax[n - 1] = A[n - 1];
           for (int i = n - 2; i >= 0; --i) {
               rightMax[i] = max(A[i+1], rightMax[i+1]);
           }
           int res = 0;
           for (int i = 1; i < n - 1; ++i) {
               leftMax = max(leftMax, A[i-1]);
               int tmp = min(leftMax, rightMax[i]);
               res += max(0, tmp - A[i]);
           }
           return res;
       }
   };