[Swift]LeetCode1292. 元素和小于等于阈值的正方形的最大边长 | Maximum Side Length of a Square with Sum Less than or Equal to Threshold
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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
给你一个大小为 m x n 的矩阵 mat 和一个整数阈值 threshold。
请你返回元素总和小于或等于阈值的正方形区域的最大边长；如果没有这样的正方形区域，则返回 0 。
示例 1：
输入：mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4
输出：2
解释：总和小于 4 的正方形的最大边长为 2，如图所示。
示例 2：
输入：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
输出：0
示例 3：
输入：mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6
输出：3
示例 4：
输入：mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184
输出：2
提示：
1 <= m, n <= 300
m == mat.length
n == mat[i].length
0 <= mat[i][j] <= 10000
0 <= threshold <= 10^5
