LeetCode—357. Count Numbers with Unique Digits

363. Max Sum of Rectangle No Larger Than K

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Total Accepted: 302 Total Submissions: 1295 Difficulty: Hard

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:

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

Credits:
Special thanks to @fujiaozhu for adding this problem and creating all test cases.

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Max Sum of Rectangle No Larger Than K思路：构建等大小的二维数组sums,sums[i][j]表示原数组直到（i,j）范围内的矩阵和

GitHub地址：https://github.com/corpsepiges/leetcode

点此进入

public class Solution {    public int maxSumSubmatrix(int[][] matrix, int k) {        int row=matrix.length;        int col=matrix[0].length;        int[][] sums=new int[row][col];        sums[0][0]=matrix[0][0];        for (int i = 1; i < row; i++) {            sums[i][0]=sums[i-1][0]+matrix[i][0];        }        for (int j = 1; j < col; j++) {            sums[0][j]=sums[0][j-1]+matrix[0][j];        }        for (int i = 1; i <row; i++) {            for (int j = 1; j < col; j++) {                sums[i][j]=sums[i-1][j]+sums[i][j-1]-sums[i-1][j-1]+matrix[i][j];            }        }        int ans=Integer.MIN_VALUE;        for (int si = 0; si < row; si++) {            for (int sj = 0; sj < col; sj++) {                for (int ei = si; ei < row; ei++) {                    for (int ej = sj; ej < col; ej++) {                        int test=0;                        if (si==0&&sj==0) {                            test=sums[ei][ej];                        }else if(si==0){                            test=sums[ei][ej]-sums[ei][sj-1];                        }else if(sj==0){                            test=sums[ei][ej]-sums[si-1][ej];                        }else{                            test=sums[ei][ej]-sums[si-1][ej]-sums[ei][sj-1]+sums[si-1][sj-1];                        }                        if (test<=k&&test>ans) {                            ans=test;                        }                    }                }            }        }        return ans;    }}