...的點,還需要判斷其所能構成的最大矩形。 public int maximalRectangle(char[][] matrix) { if(matrix.length==0 || matrix[0].length==0) return 0; int row = matrix.length; int column = matrix[0]....
...n = 2, the maximum dp[i] = 2 appeared twice, indicating that there are two maximal squares. Solution public class Solution { public int maxSquare(int[][] matrix) { int mLen = 0; ...
Maximal Square Given a 2D binary matrix filled with 0s and 1s, find the largest square containing all 1s and return its area. For example, given the following matrix: 1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 ...
... 1 0 0 1 0 return 4 // O(mn) space public class Solution { public int maximalSquare(char[][] matrix) { if(matrix == null || matrix.length == 0) return 0; int m = matrix.length,...
...j] + 1)(橫長) * height[j]就是可能的最大的矩形了。 public int maximalRectangle(char[][] matrix) { if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { return 0; } ...
...][0] = 0; //Result: f[matrix.length][matrix[0].length]; public int maximalSquare(char[][] matrix) { if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { ...
...3 5 5 1 5 3 5 5 1 5 3 5 5 1 5 5 4 5 public class Solution { public int maximalRectangle(char[][] matrix) { int m = matrix.length; if(matrix == null || m == 0) return 0; ...
... ,7], [3 ,7 ,8], [4 ,8 ,9], ] return 5 Challenge O(k log n), n is the maximal number in width and height. Note Solution I. Muggle (95% ac, last case exceeded time limit) public class Solution ...
...s.erase(bucket); } } /** Returns one of the keys with maximal value. */ string getMaxKey() { return buckets.empty() ? : *(buckets.rbegin()->keys.begin()); } ...
...s.erase(bucket); } } /** Returns one of the keys with maximal value. */ string getMaxKey() { return buckets.empty() ? : *(buckets.rbegin()->keys.begin()); } ...
...anteed to be a non-empty string.GetMaxKey() - Returns one of the keys with maximal value. If no element exists, return an empty string .GetMinKey() - Returns one of the keys with minimal value. If ...
... return root; } // 返回最大值節點 public TreeNode maximal(TreeNode root){ if(root ==null){ return null; } while(root.right!=null){ root = ...
...th.max(currMax, max); } } return max; } } Maximal Rectangle Given a 2D binary matrix filled with 0s and 1s, find the largest rectangle containing all ones and retur...
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