42. Trapping Rain Water

42. Trapping Rain Water

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!

解析

• 找到最高水位，然后记录左边最大水位，低于最大水位就累加，从最右边记录最高水位，低于水位就累加！

// add 42. Trapping Rain Water
class Solution_42 {
public:

	// 本来自己想用总面积-黑色块的面积，但是总面积不容易求得
	// 思路1：找到最高的柱子，分左右两边处理 
	int trap(vector<int>& height) {

		if (height.size()<=0)
		{
			return 0;
		}
		int max_index = 0;
		int max_height = height[0];
		for (int i = 1; i < height.size();i++)
		{
			if (height[i]>max_height)
			{
				max_height = height[i];
				max_index = i;
			}
		}

		int sum = 0;
		int max_left = 0;
		for (int i = 0; i < max_index;i++)
		{
			if (height[i]>max_left)
			{
				max_left = height[i];
			}
			else
			{
				sum += (max_left-height[i]);
			}
		}

		int max_right = 0;
		for (int i = height.size() - 1; i >max_index; i--)
		{
			if (height[i]>max_right)
			{
				max_right = height[i];
			}
			else
			{
				sum += (max_right-height[i]);
			}
		}
		return sum;
	}

	int trap(int A[], int n) {
		vector<int > vec(A,A+n);
		
		return trap(vec);

	}
};

题目来源

• 42. Trapping Rain Water