leetcode;Trapping Rain Water

最新推荐文章于 2020-08-24 17:46:40 发布

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本文介绍了一种计算在下雨后，给定高度数组所代表的地形中能积聚多少雨水的算法。通过示例解释了如何计算给定数组[0,1,0,2,1,0,1,3,2,1,2,1]中的雨水量为6单位。

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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(vector<int>& height) {
        
        if (height.size() == 0)
            return 0;
            
        int pivot = 0;
        int maxValue = 0;
        for (int i=0; i<height.size(); i++)
        {
            if (height[i] > maxValue)
            {
                maxValue = height[i];
                pivot = i;
            }
        }
            
        int curSum = 0;
        int curMax = height[0];
        for (int i=1; i<pivot; i++)
        {
            if (curMax > height[i])
                curSum += (curMax-height[i]);
            else if (curMax < height[i])
                curMax = height[i];
        }
        
        curMax = height[height.size()-1];
        for (int i=height.size()-2; i>pivot; i--)
        {
            if (curMax > height[i])
                curSum += (curMax-height[i]);
            else if (curMax < height[i])
                curMax = height[i];
        }
        
        return curSum;
    }
};