[leetcode] 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(n<3)
return 0;
int start,end,sum;
start=0,end=0,sum=0;
for( ; end<n ; ){
if( A[start]<A[end] && end>start )
start=end;
else{
sum+=A[end]-A[start];
end++;
}
}
return sum;
}
};

上面这个代码是不行的，因为没有考虑到A[end]可能不会再比start大了，所以我就分开来，再递归调用一次

class Solution {
public:
int trap(int A[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(n<3)
return 0;
int start,end,sum,sumt;
start=end=sum=sumt=0;
for( ; end<n ; ){
if(A[start]<=A[end] && end>start){
sum+=sumt;
sumt=0;
start=end;
}
else{
sumt+=A[start]-A[end];
end++;
if(end==n && start!=n){
int count=end-start;
int *p=new int[count];
for(int i=0 ; i<count ; i++){
p[i]=A[--end];
}
sumt=trap(p,count);
return sumt+sum;
}
}
}
return sum;
}
};

当然还有一种比较容易理解的方法：
class Solution {
public:
int trap(int A[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> left(n);
int maxHeight = 0;
for(int i = 0; i < n; i++)
{
left[i] = maxHeight;
maxHeight = max(maxHeight, A[i]);
}

vector<int> right(n);
maxHeight = 0;
for(int i = n - 1; i >= 0; i--)
{
right[i] = maxHeight;
maxHeight = max(maxHeight, A[i]);
}

int water = 0;
for(int i = 0; i < n; i++)
{
int height = min(left[i], right[i]) - A[i];
if (height < 0)
height = 0;
water += height;
}

return water;
}
};