LeetCode Trapping Rain Water

coobj2008

于 2014-06-04 20:49:14 发布

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本文介绍了一种计算在下雨后，给定高度数组所形成的地形中能容纳多少雨水的算法。通过遍历数组并计算每个位置左右两侧最高点的高度差与当前位置高度之差，最终得出总积水量。

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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.

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!

class Solution {
public:
	int trap(int A[], int n) {
		if (n <= 2)
			return 0;
		int maxH = 0;
		//left[i]表示i左边最高的值，right[i]表示i右边最高的值
		int *left= new int[n], *right = new int[n];
		for (int i = 0; i < n; i++) {
			maxH = max(maxH, A[i]);
			left[i] = maxH;
		}
		maxH = 0;
		for (int i = n - 1; i >= 0; i--) {
			maxH = max(maxH, A[i]);
			right[i] = maxH;
		}
		int ans = 0;
		for (int i = 0; i < n; i++)
			ans += (min(left[i], right[i]) - A[i]);
		delete[] left;
		delete[] right;
		return ans;
	}
};