leetcode_1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold

题目链接

Given a m x n matrix mat and an integer threshold. Return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square.

Example 1:

Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4

Output: 2

Explanation: The maximum side length of square with sum less than 4 is 2 as shown.

Example 2:

Input: mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1

Output: 0

Example 3:

Input: mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6

Output: 3

Example 4:

Input: mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184

Output: 2

Constraints:

1 <= m, n <= 300
m == mat.length
n == mat[i].length
0 <= mat[i][j] <= 10000
0 <= threshold <= 10^5

给定一个矩阵，找出一个最大正方形的边长，这个正方形中元素和要小于threshold。

解法：

二维前缀和，sum[i][j]表示从[0][0]到[i][j]的矩形的元素和，有了这个sum后，就可以用 O(row * col * min(row,col)) 的时间复杂度遍历所有的正方形。

class Solution {

public:

    int maxSideLength(vector<vector<int>>& mat, int threshold) {

        int row = mat.size(), col = mat[].size();

        vector<vector<int>> sum(row+, vector<int>(col+, ));

        for(int i=; i<=row; i++)

            for(int j=; j<=col; j++)

                sum[i][j] = sum[i-][j]+sum[i][j-]-sum[i-][j-]+mat[i-][j-];

        int ret = ;

        for(int i=; i<=row; i++)

            for(int j=; j<=col; j++){

                for(int k=; k<=min(i, j); k++){

                    int temp = sum[i][j]-sum[i-k][j]-sum[i][j-k]+sum[i-k][j-k];

                    if(temp <= threshold)

                        ret = max(ret, k);

            }

        }

        return ret;

    }

};

巴特西

leetcode_1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold_[二维前缀和]

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