[LeetCode] Trapping Rain Water

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!

Solution:

基本思路是独立考虑每一个单位能容下的谁。比如上例中，坐标x~[2, 3]能装1个单位的水，[5, 6]能装2个单位的水，而能装的水的体积与该位置向左最高值LeftHighest，和向右最高值RightHighest相关。volume = max(0, min(LeftHighest, RightHighest))，有了这个认识，体积就很容易算出了。

class Solution {
public:
int *leftHigh, *rightHigh;

int trap(int A[], int n) {
if(n < 3) return 0;

int volume = 0;
leftHigh = new int
, rightHigh = new int
;

int left = 0, right = 0;
for(int i = 0;i < n;i++)
{
leftHigh[i] = left;
if(A[i] > left) left = A[i];
rightHigh[n - 1 - i] = right;
if(A[n - 1 - i] > right) right = A[n - 1 - i];
}
for(int i = 0;i < n;i++)
volume += max(0, min(leftHigh[i], rightHigh[i]) - A[i]);

return volume;
}
};