[Swift]LeetCode363. 矩形区域不超过 K 的最大数值和 | Max Sum
发布时间:2020-12-14 05:05:46 所属栏目:百科 来源:网络整理
导读: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: Input: matrix = [[1,1],[0,-2,3]],k = 2 Output: 2 Explanation:?Because the sum of rect
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: Input: matrix = [[1,1],[0,-2,3]],k = 2 Output: 2 Explanation:?Because the sum of rectangle is 2,? and 2 is the max number no larger than k (k = 2).[[0,[-2,3]]
Note:
给定一个非空二维矩阵?matrix?和一个整数?k,找到这个矩阵内部不大于?k的最大矩形和。 示例: 输入: matrix = [[1,k = 2 输出: 2 解释:?矩形区域??的数值和是 2,且 2 是不超过 k 的最大数字(k = 2)。 [[0,3]] 说明:
13108ms 1 class Solution { 2 func maxSumSubmatrix(_ matrix: [[Int]],_ k: Int) -> Int { 3 if matrix.isEmpty || matrix[0].isEmpty {return 0} 4 var m:Int = matrix.count 5 var n:Int = matrix[0].count 6 var res:Int = Int.min 7 var sum:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n),count:m) 8 for i in 0..<m 9 { 10 for j in 0..<n 11 { 12 var t:Int = matrix[i][j] 13 if i > 0 {t += sum[i - 1][j]} 14 if j > 0 {t += sum[i][j - 1]} 15 if i > 0 && j > 0 {t -= sum[i - 1][j - 1]} 16 sum[i][j] = t 17 for r in 0...i 18 { 19 for c in 0...j 20 { 21 var d:Int = sum[i][j] 22 if r > 0 {d -= sum[r - 1][j]} 23 if c > 0 {d -= sum[i][c - 1]} 24 if r > 0 && c > 0 {d += sum[r - 1][c - 1]} 25 if d <= k {res = max(res,d)} 26 } 27 } 28 } 29 } 30 return res 31 } 32 } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |