LeetCode : Trapping Rain Water (java)

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：先找到最高点，从两侧到最高点只增不减，发现减少的，即需要“加水”填平，由此可以计算出盛水量。

public class Solution {
public int trap( int[] height )
{
if ( height == null || height.length <= 2 )
{
return(0);
}

int area = 0;
/* 找到最高点 */
int max = height[0], maxIndex = 0;
for ( int i = 1; i < height.length; ++i )
{
if ( height[i] > max )
{
max		= height[i];
maxIndex	= i;
}
}

/* 左侧遍历 */
int left = height[0];
for ( int i = 1; i < maxIndex; ++i )
{
if ( left < height[i] )
{
left = height[i];
}else{
area += left - height[i];
}
}

/* 右侧遍历 */
int right = height[height.length - 1];
for ( int i = height.length - 1; i > maxIndex; --i )
{
if ( right < height[i] )
{
right = height[i];
}else{
area += right - height[i];
}
}

return(area);
}
}