Leetcode: Trapping Rain Water

最新推荐文章于 2018-06-27 09:33:25 发布

carlosten

最新推荐文章于 2018-06-27 09:33:25 发布

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分类专栏： Leetcode 文章标签： algorithm 面试

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本文介绍了一种算法，用于计算给定高度数组所代表的地形中，在雨后能够储存的雨水量。通过定义左右边界并计算每个位置可能达到的最大高度，算法能够准确估算出储水量。

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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!

c++:

class Solution {
public:
    int trap(int A[], int n) {
        if(n == 0)
        {
            return 0;
        }
        int l[n];
        int r[n];
        int i;
        int h = 0;
        for(i = 0;i < n;i ++)
        {
            if(A[i] > h)
            {
                h = A[i];
            }
            l[i] = h;
        }
        h = 0;
        for(i = n-1;i >= 0;i --)
        {
            if(A[i] > h)
            {
                h = A[i];
            }
            r[i] = h;
        }
        int sum = 0;
        for(i = 0;i < n;i ++)
        {
            sum = sum + min(l[i],r[i]) - A[i];
        }
        return sum;
    }
};