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：关键搞清楚长短板对水量的影响，此题可以认为由最高的水柱隔开左右两个大池子。

（1）对左面的大池子计算水量：

从左向右先找到第一个不为0的柱子；然后以这个柱子为基准，将其后比它小的柱子累计起来，求水量和，直到它成为短板，即出现比基准柱子更高的柱子。

更新此时最高的柱子为基准。

一直重复上述过程，直到最高的水柱结束左面大池子的水量统计；

（2）从右边开始找水池，并与左面相同，依次累计水量和，直到最高水柱，结束；

左右两边的水量和即是结果。

class Solution
{
public:
int trap(vector<int>& height)
{
int n = height.size();
int current = 0;
int cur_index = 0;
int i = 0;
for(i = 0; i<n; i++)
{
if(height[i] != 0)
{
current = height[i];
cur_index = i;
break;
}
}
int sum = 0;

int cur_sum = 0;
for(i = cur_index; i<n; i++)
{
if(height[i] < current)
{
cur_sum = cur_sum + current - height[i];
}
else if(height[i] >= current)
{
cur_index = i;
current = height[i];
sum = sum + cur_sum;
cur_sum = 0;
}

}
if(cur_index <n)
{

int daoxu = cur_index;
cur_index = n-1;
current = height[cur_index];
for(i = n-2; i>= daoxu; i--)
{
if(height[i] >= current)
{
cur_index = i;
current = height[cur_index];
continue;
}
else
{
break;
}
}
cur_sum = 0;
for(i =cur_index; i>=daoxu; i-- )
{
if(height[i] < current)
{
cur_sum = cur_sum + current - height[i];
}
else if(height[i] >= current)
{
cur_index = i;
current = height[i];
sum = sum + cur_sum;
cur_sum = 0;
}
}

}
return sum;
}
};