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)。

1 class Solution { 2 public: 3     int trap(int A[], int n) { 4         int maxHeight = 0; 5         vector
     
       leftMostHeight(n, 0); 6         for (int i = 0; i < n; ++i) { 7             leftMostHeight[i] = maxHeight; 8             maxHeight = (maxHeight > A[i]) ? maxHeight : A[i]; 9         }10         maxHeight = 0;11         vector
      
        rightMostHeight(n, 0);12         for (int i = n - 1; i >= 0; --i) {13             rightMostHeight[i] = maxHeight;14             maxHeight = (maxHeight > A[i]) ? maxHeight : A[i];15         }16         int water = 0, height;17         for (int i = 0; i < n; ++i) {18             height = min(leftMostHeight[i], rightMostHeight[i]) - A[i];19             water += (height > 0) ? height: 0;20         }21         return water;22     }23 };
      
     

 下面是扫一遍的代码：

1 class Solution { 2 public: 3     int trap(int A[], int n) { 4         int left = 0, right = n - 1; 5         int height = 0, res = 0; 6         while (left < right) { 7             if (A[left] < A[right]) { 8                 height = max(height, A[left]); 9                 res += height - A[left];10                 ++left;11             } else {12                 height = max(height, A[right]);13                 res += height - A[right];14                 --right;15             }16         }17         return res;18     }19 };