连续数组最大和 (二维数组)

二维矩阵最大子矩阵求和

最新推荐文章于 2021-12-05 21:43:17 发布

转载最新推荐文章于 2021-12-05 21:43:17 发布 · 547 阅读

本文介绍了一种算法，用于寻找二维整数矩阵中具有最大和的子矩阵，并通过具体示例展示了如何使用该算法。算法首先计算矩阵中所有可能的矩形区域的累积和，然后遍历这些累积和来确定最大子矩阵的具体位置和总和。

public void getMaxSubArray(int[][]array){

        int row = array.length;
        int width = array[0].length;
        int[][] fe = new int[row+1][width+1];
        for(int i=1;i<=row;i++){
            for(int j=1;j<=width;j++){
                fe[i][j] = fe[i][j-1] + fe[i-1][j] - fe[i-1][j-1] + array[i-1][j-1];
            }
        }

        int finalMax =0,finalLow = -1,finalHigh = -1,finalB = -1,finalE = -1;

        for(int low=0;low<row;low++){
            for(int high=low+1;high<=row;high++){
                int max = 0,sum = 0,b=-1,bt=-1,e=-1;
                for(int j=1;j<=width;j++){
                    sum += getValue(fe,low,high,j);

                    if(sum>max){
                        max = sum;
                        e = j - 1;
                        b = bt;
                    }
                    if(sum<0){
                        sum = 0;
                        bt = j - 1;
                    }
                }
                if(max>finalMax){
                    finalMax = max;
                    finalB = b;
                    finalE = e;
                    finalHigh = high-1;
                    finalLow = low-1;
                }
            }
        }

        for(int i=finalLow+1;i<=finalHigh;i++){
            for(int j=finalB+1;j<=finalE;j++){
                System.out.print(array[i][j]+ ",");
            }
            System.out.println("");
        }
    }

    public int getValue(int[][]fe,int low,int high,int col){
        return fe[high][col] - fe[low][col] - fe[high][col-1] + fe[low][col-1];
    }

    public static void main(String[]args){
        MaxSubArrayII arrayII = new MaxSubArrayII();
        arrayII.getMaxSubArray(new int[][]{
                {1,-10,-11},
                {4,-95,6},
                {7,8,9}
        });
    }

转载：https://blog.youkuaiyun.com/zhumingyuan111/article/details/80193595