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!
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;
}
}; |