class Solution { public: int max = INT_MIN; int maxPathSum(TreeNode *root) { if (root == NULL) return 0; search(root); return max; } int search(TreeNode *root) { if (root == NULL) return 0; int left_max = search(root->left); int right_max = search(root->right); int ...
78 -- 9:48 App LeetCode刷题日记 Day 23 Part 2 - Search in Rotated Sorted Array 154 -- 3:19 App LeetCode刷题日记 Day 40 Part 1 - Matrix Diagonal Sum 89 -- 5:01 App LeetCode刷题日记 Day 6 Part 2 - Binary Tree Level Order Traversal浏览...
代码如下: 1publicclassSolution {2intmax =Integer.MIN_VALUE;3publicintmaxPathSum(TreeNode root) {4if(root ==null)return0;5dfs(root);6returnmax;7}8privateintdfs(TreeNode root){9if(root ==null)return0;10intleft =dfs(root.left);11intright =dfs(root.right);12max = Math.max(max,Math...
(2) Matrix Rounding Problem (矩阵取整问题) 定义矩阵 M_{p\times q} ,对矩阵进行取整操作,要求是对矩阵的每个元素都要取整, \lfloor M_{ij} \rfloor 或\lceil M_{ij} \rceil 同时要求取整后的矩阵的每行的元素的和等于原矩阵对应行元素的和的取整,同样对列也有类似的要求。如下所示是一个3乘3的...
Leetcode: Binary TreeMaximumPathSum 难度:75. 参见了他人的思路,这道题是求树的路径和的题目,不过和平常不同的是这里的路径不仅可以从根到某一个结点,而且路径可以从左子树某一个结点,然后到达右子树的结点,就像题目中所说的可以起始和终结于任何结点。函数的返回值定义为以自己为根的一条从根到叶子结点的最长...
329-longest-increasing-path-in-a-matrix 33-search-in-rotated-sorted-array 330-patching-array 3307-find-the-maximum-sum-of-node-values README.md find-the-maximum-sum-of-node-values.py 332-reconstruct-itinerary 3332-minimum-operations-to-exceed-threshold-value-ii 334-increasing-triplet-subsequence...
0054-spiral-matrix.cpp 0055-jump-game.cpp 0056-merge-intervals.cpp 0057-insert-interval.cpp 0058-length-of-last-word.cpp 0061-rotate-list.cpp 0062-unique-paths.cpp 0063-unique-paths-ii.cpp 0064-minimum-path-sum.cpp 0066-plus-one.cpp 0067-Add-Binary.cpp 0067-add-binary.cpp 0069-sqrt(x)...
2, the aspiration criterion is applied if a neighbor solution is listed as a forbidden move in the Tabu Matrix but has a better objective function value compared to the best value found so far. 4.4 Simheuristic with intensive simulation In addition to incorporating MCS during the neighborhood ...
As cells are fixed in the solid matrix, there exists no washout here. Or the cells are separated by the membrane from the flow stream. The substrate is fed to the flow stream that can enter the cell chamber by diffusion and/or pressure-ultrafiltration (UF) swing method [61, 62]. The ...
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