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Trapping Rain Water
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!
class Solution {
public:
int trap(int A[], int n)
{
int count=0;
if(n<=2) return count;
int h[n];h[0]=A[0];
for(int i=1;i<n;i++)
if(A[i]>h[i-1]) h[i]=A[i];
else h[i]=h[i-1];
h[n-1]=A[n-1];
for(int i=n-2;i>=0;i--)
{
if(h[i]<=h[i+1]) break;
if(A[i]>h[i+1]) h[i]=A[i];
else h[i]=h[i+1];
}
for(int i=0;i<n;i++)
count=count+h[i]-A[i];
return count;
}
};
public:
int trap(int A[], int n)
{
int count=0;
if(n<=2) return count;
int h[n];h[0]=A[0];
for(int i=1;i<n;i++)
if(A[i]>h[i-1]) h[i]=A[i];
else h[i]=h[i-1];
h[n-1]=A[n-1];
for(int i=n-2;i>=0;i--)
{
if(h[i]<=h[i+1]) break;
if(A[i]>h[i+1]) h[i]=A[i];
else h[i]=h[i+1];
}
for(int i=0;i<n;i++)
count=count+h[i]-A[i];
return count;
}
};
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