[leetcode] Trapping Rain Water

最新推荐文章于 2020-08-24 17:46:40 发布

和道一文字JC

最新推荐文章于 2020-08-24 17:46:40 发布

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分类专栏：算法题

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本文探讨了一种计算二维地形中雨水滞留量的算法。通过分析给定的高度数组，采用双指针技术并结合最大高度预处理策略，实现了高效求解。此外，提供了多种实现方案对比。

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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(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(n<3)
            return 0;
        int start,end,sum;
        start=0,end=0,sum=0;
        for( ; end<n ; ){
            if( A[start]<A[end] && end>start )
                start=end;
            else{
                sum+=A[end]-A[start];
                end++;
            }
        }
        return sum;
    }
};

上面这个代码是不行的，因为没有考虑到A[end]可能不会再比start大了，所以我就分开来，再递归调用一次

class Solution {
public:
    int trap(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(n<3)
            return 0;
        int start,end,sum,sumt;
        start=end=sum=sumt=0;
        for( ; end<n ; ){
            if(A[start]<=A[end] && end>start){
                sum+=sumt;
                sumt=0;
                start=end;
            }
            else{
                sumt+=A[start]-A[end];
                end++;
                if(end==n && start!=n){
                    int count=end-start;
                    int *p=new int[count];
                    for(int i=0 ; i<count ; i++){
                        p[i]=A[--end];
                    }
                    sumt=trap(p,count);
                    return sumt+sum;
                }
            }
        }
        return sum;
    }
};

当然还有一种比较容易理解的方法：

class Solution {
public:
    int trap(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> left(n);
        int maxHeight = 0;
        for(int i = 0; i < n; i++)
        {
            left[i] = maxHeight;
            maxHeight = max(maxHeight, A[i]);
        }
        
        vector<int> right(n);
        maxHeight = 0;
        for(int i = n - 1; i >= 0; i--)
        {
            right[i] = maxHeight;
            maxHeight = max(maxHeight, A[i]);
        }
        
        int water = 0;
        for(int i = 0; i < n; i++)
        {
            int height = min(left[i], right[i]) - A[i];
            if (height < 0)
                height = 0;
            water += height;
        }
        
        return water;
    }
};