Trapping Rain Water

最新推荐文章于 2024-09-17 16:23:14 发布

Nereus_Li

最新推荐文章于 2024-09-17 16:23:14 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/li_chihang/article/details/44836427

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本文介绍了一种算法，用于计算给定高度数组所代表的地形中，在雨后能够储存的雨水量。通过使用左右最高墙的概念，算法能够高效地计算出储存的雨水总量。

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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!

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

int trap(int A[], int n) {
	if (n == 0)
		return 0;
	vector<int>Lefthigh(n, 0);
	vector<int>Righthigh(n, 0);
	Lefthigh[0] = A[0];
	Righthigh[n - 1] = A[n - 1];

	for (int i = 1; i < n;++i)
		Lefthigh[i] = A[i]>Lefthigh[i - 1] ? A[i] : Lefthigh[i - 1];
	for (int i = n - 2; i >= 0; --i)
		Righthigh[i] = A[i] > Righthigh[i + 1] ? A[i] : Righthigh[i + 1];
	int sum = 0;
	for (int i = 0; i != n;++i)
	{
		int tmp = min(Lefthigh[i], Righthigh[i]) - A[i]>0 ? min(Lefthigh[i], Righthigh[i]) - A[i]:0;
		sum += tmp;
	}
	return sum;
}