题目:
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!
思路:
首先从两边向中间找到最高的那一根柱子。
然后从两边分别向中间移动并灌水。我们用floor来表示当前水平面,如果水平面比柱子高,那么这个区域可以灌水;如果水平面比柱子低,那么向里的水平面可以增加。
代码:
class Solution {
public:
int trap(int A[], int n) {
if(n<2)
return 0;
int highest=A[0];
int hindex = 0;
for(int i=1;i<n;i++)
{
if(A[i]>highest)
{
highest = A[i];
hindex = i;
}
}
int water=0;
int floor=0;;
for(int i=0;i<hindex;i++)
{
if(floor>=A[i])
{
water+=floor-A[i];
}
else
{
floor=A[i];
}
}
floor=0;;
for(int i=n-1;i>hindex;i--)
{
if(floor>=A[i])
{
water+=floor-A[i];
}
else
{
floor=A[i];
}
}
return water;
}
};