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!

分析

对于每个柱子，找到其左右两边最高的柱子，该柱子能容纳的面积就是min(max_left,

max_right) - height。所以，

1. 从左往右扫描一遍，对于每个柱子，求取左边最大值；

2. 从右往左扫描一遍，对于每个柱子，求最大右值；

3. 再扫描一遍，把每个柱子的面积并累加。

也可以，

1. 扫描一遍，找到最高的柱子，这个柱子将数组分为两半；

2. 处理左边一半；

3. 处理右边一半。

代码：

class Solution {
public:
int trap(int A[], int n)
{
if (n==0)
{
return 0;
}
n=n-1;
int maxNumber=FindMax(A,0,n);
int l=leftSolution(A,maxNumber);
int r=rightSolution(A,maxNumber,n);
return l+r;

}
public:
int rightSolution(int A[],int r,int n)
{

if (r==n)
{
return 0;
}
int maxNumber=FindMax(A,r+1,n);
int sumNumber=0;
for (int i = maxNumber-1; i >=r+1; i--)
{
sumNumber+=A[i];

}
int sum=A[maxNumber]*(maxNumber-r-1)-sumNumber;
return rightSolution(A,maxNumber,n)+sum;
}
public:
int leftSolution(int A[],int l)
{
if (l==0)
{
return 0;
}

int maxNumber=FindMax(A,0,l-1);
int sumNumber=0;
for (int i = maxNumber+1; i <=l-1; i++)
{
sumNumber+=A[i];
}
int sum=A[maxNumber]*(l-maxNumber-1)-sumNumber;
return leftSolution(A,maxNumber)+sum;

}
public:
int FindMax(int A[],int l,int h)
{
int maxNum=l;
for (int i = l; i < h+1; i++)
{
if (A[maxNum]<A[i])
{
maxNum=i;
}
}
return maxNum;

}
};