Max Sum of Rectangle No Larger Than K

小灰灰52309

于 2017-05-12 14:50:53 发布

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分类专栏： leetcode ACM 文章标签： Max Sum of Rectangle

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本文介绍了一种解决二维矩阵中寻找最大和的矩形子矩阵的问题，该子矩阵的和不超过给定值k的方法。通过累加求和的方式，算法能够有效地找到满足条件的最大和。

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Max Sum of Rectangle No Larger Than K

Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k.

Example:

Given matrix = [
  [1,  0, 1],
  [0, -2, 3]
]
k = 2

The answer is 2. Because the sum of rectangle [[0, 1], [-2, 3]] is 2 and 2 is the max number no larger than k (k = 2).

Note:

The rectangle inside the matrix must have an area > 0.
What if the number of rows is much larger than the number of columns?

解析：

使用累加求和来求每个子矩阵的和。

代码：

class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        
        int rows=matrix.size();
        int cols=matrix[0].size();
        vector<vector<int>> Sum(rows+1,vector<int>(cols+1,0));
        int ans=INT_MIN;
        for (int i=1; i<=rows; i++)
        {
            for (int j=1; j<=cols; j++)
            {
                Sum[i][j]=Sum[i-1][j]+Sum[i][j-1]-Sum[i-1][j-1]+matrix[i-1][j-1];
                
                
                for (int r=0; r<i; r++)
                {
                    for (int c=0; c<j; c++)
                    {
                        int temp=Sum[i][j]-Sum[r][j]-Sum[i][c]+Sum[r][c];
                        if (temp<=k&&temp>ans)
                        ans=temp;
                    }
                }
            }
        }
        return ans;
        
    }
};