Leetcode之Trapping Rain Water

最新推荐文章于 2020-07-18 01:52:58 发布

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本文介绍了一种计算雨后地形积水体积的算法。通过分析地形高度数据，利用两边向中间逼近的方法，高效准确地计算出地形能容纳的雨水量。以数组[0,1,0,2,1,0,1,3,2,1,2,1]为例，详细展示了算法的实现过程，最终计算出积水总量为6单位。

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题目：

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.

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!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

代码：

class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
	if (n <= 2)return 0;
	int max = -1, maxInd = 0;
	int i = 0;
	for (; i < n; i++) {
		if (height[i] > max) {
			max = height[i];
			maxInd = i;
		}
	}

	int area = 0, root = height[0];
	for (i = 0; i < maxInd; i++) {
		if (root < height[i]) {
			root = height[i];
		}
		else area += root - height[i];
	}

	for (i = n - 1, root = height[n - 1]; i > maxInd; i--) {
		if (root < height[i]) {
			root = height[i];
		}
		else area += root - height[i];
	}

	return area;
    }
};

注：

思路————看成一个塔形，从两边往中间！

主要还是思路！