[Leetcode] [Trapping Rain Water 2012-03-10]

Given n non-negative integers representing an elevation map where the width of each bar are 1, compute how much WA ter It is the 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 the Rain Water (blue section) is being trapped. Thanks Marcos for contributing this image!

The key problem is Get B[i] and C[i], where b[i] means the max forward, and c[i] means Max backward.

Sum + = Min (B[i], c[i])-a[i] >0? Min (B[i], c[i])-a[i]:0

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class Solution { public:     int trap(int A[], int n) {         int* B = new int[n];         int sum = 0;         int max= 0;         for(int i = 0 ;i<n;i++)         {             B[i]=max;             if(max < A[i])             {                 max=A[i];                }         }         max =0 ;         for(int i=n-1; i>=0 ;i--)         {             int h = B[i];             if(B[i] > max)             {                 h = max;             }             if(A[i] < h)             {                 sum += h-A[i];             }                         if(max < A[i])             {                 max=A[i];                }         }         delete[] B;         return sum;                     } };