042 - Trapping Rain Water

最新推荐文章于 2022-03-06 09:58:25 发布

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本文介绍了一种计算二维数组表示的地面上雨后积水总量的方法。通过寻找左侧最高点和右侧最高点来确定每个位置上可能积累的最大水量，并最终计算总的积水体积。

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!

int findleft(int *height, int size, int loc)
{
	if (height[loc - 1] <= height[loc]) return -1;
	else return loc - 1;
}

int findright(int *height, int size, int loc)
{
	int i, right = -1;
	for (i = loc + 1; i < size; i++) {
		if ((height[i] > height[loc]))
			if (right < 0)
				right = i;
			else if (height[i] > height[right])
				right = i;
		if (height[right] >= height[loc - 1]) break;
	}
	return right;
}

int trap(int* height, int heightSize) 
{
	int i, j, left, right;
	int ret = 0;
	for (i = 1; i < heightSize - 1; i++) {
		left = findleft(height, heightSize, i);
		right = findright(height, heightSize, i);
		if (left >= 0 && right > 0) {
			int h = height[left] < height[right]? height[left]:height[right];
			for (j = left + 1; j < right; j++) {
				ret += h - height[j];
			}	
			i = right;
		}
	}

	return ret;
}