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