继上一节学习了和最大连续子串，推广到和最大连续子矩阵，以二维为例，子矩阵是矩阵行和列的组合，参考：

代码如下：

package programming;/**最大和的连续子矩阵 * 将连续子矩阵看做是一维连续子串，然后利用和最大连续子串方法，求出连续子串，进而求出子矩阵 * @author ywf * */public class MaxSumSubMatrix {    /**     * @param args     */    public static void main(String[] args) {        int[][] A =  {
   {0, -2,  -7,  0},                  {
    9 , 2 , -6,  2},                  {
    -4 , 1 ,-4 , 1},                  {
    -1 , 8 ,  0 ,-2}};        int[] temp = new int[A[0].length];        int subMaxSum = Integer.MIN_VALUE;        int colStart = -1;        int colEnd = -1;        int rowStart = -1;        int rowEnd = -1;        //矩阵压缩        for(int i = 0 ;i
     
      subMaxSum){                    subMaxSum = tempMax[2];                    colStart = tempMax[0];                    colEnd = tempMax[1];                    rowStart = i;                    rowEnd = j;                }                temp = new int[A[0].length];            }                    }        //输出和最大子矩阵        for(int i = rowStart;i<=rowEnd;i++){            for(int j = colStart;j<=colEnd;j++){                System.out.print(A[i][j]+" ");            }            System.out.println();        }        System.out.println("sum:"+subMaxSum);    }    public static int[] getResult4(int[] array) {        int[] result = new int[3];        int rmax = Integer.MIN_VALUE;//当前最大的和        int sum = Integer.MIN_VALUE;//当前的和        int start = -1;//最大子串的起始下标        int end = -1;//最大子串的结束下标        for (int i = 0; i < array.length; i++) {            if (sum > 0) {                sum += array[i];            } else {                sum = array[i];                if(sum>rmax){                    start = i;                }                            }            if (sum > rmax) {                rmax = sum;                end = i;            }        }        result[0] = start;        result[1] =end ;        result[2] = rmax;        return result;    }}
     

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