【LeetCode】Trapping Rain Water 2013年美团网校园招聘研发工程师笔试题

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!

Discuss
java code : lhs 对应能装水的台阶的左边界，rhs对应右边界，在这个边界确定的范围内，对每一个A[i] ，其能装到的水的大小为 min{左边最高值 ，右边界最高值} - A[i] if ( 该值 > A[i]) .我们预处理出这样一个左边界的最高值数组

lheight[] ，然后从右边扫过来即可。算法的复杂度是O(n)。

public class Solution {
public int trap(int[] A) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int length = A.length;
if(length <= 2)
return 0;
int lhs = 0, rhs = length - 1;
while((lhs + 1) < length)
{
if(A[lhs + 1] >= A[lhs])
lhs++;
else break;
}
while(rhs > 0)
{
if(A[rhs - 1] >= A[rhs])
rhs--;
else break;
}
int[] lheight = new int[length];
int maxheight = A[lhs];
for(int i = lhs + 1; i < rhs; i++)
{
lheight[i] = maxheight;
if(A[i] > maxheight)
maxheight = A[i];
}
maxheight = A[rhs];
int res = 0;
for(int i = rhs - 1; i > lhs; i--)
{
int minheight = Math.min(maxheight, lheight[i]);
if(minheight > A[i])
res += minheight - A[i];
if(A[i] > maxheight)
maxheight = A[i];
}
lheight = null;
return res;

}
}