[LeetCode] 118: Trapping Rain Water

最新推荐文章于 2024-09-02 03:43:19 发布

1.02^365的成长裂变

最新推荐文章于 2024-09-02 03:43:19 发布

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本文针对一个特定的问题——如何计算地形图中能留住多少雨水——提供了一种有效的算法解决方案。通过使用动态规划的方法预处理出每个位置左侧和右侧的最高点，进而计算出每个位置能够保留的雨水量。

[Problem]

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.

[LeetCode] 118: Trapping Rain Water - coder007 - Coder007的博客

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]

class Solution {
public:
	int trap(int A[], int n) {
		// Note: The Solution object is instantiated only once and is reused by each test case.
		// initial
		int leftBoder[n], rightBoder[n];

		// dp
		int L = 0, R = 0;
		for(int i = 0; i < n; ++i){
			if(i == 0){
				L = A[i];
				R = A[n-1-i];
			}
			else{
				L = L > A[i] ? L : A[i];
				R = R > A[n-1-i] ? R : A[n-1-i];
			}
			leftBoder[i] = L;
			rightBoder[n-1-i] = R;
		}

		// get result
		int res = 0, boder;
		for(int i = 0; i < n; ++i){
			boder = min(leftBoder[i], rightBoder[i]);
			if(boder > A[i])res += boder - A[i];
		}
		return res;
	}
};