algorithm

问题 I was trying to solve this problem, and I gave up, and found the solution below, although I do not understand how the solution works, or why it works. Any in-depth solution would be deeply appreciated. Question: Given s1 , s2 , s3 , find whether s3 is formed by the interleaving of s1 and s2 . For example, Given: s1 = "aabcc" s2 = "dbbca" When s3 = "aadbbcbcac" , return true. When s3 = "aadbbbaccc" , return false. Solution: public static boolean isInterleave(String s1, String s2, String s3) {

问题 It seems easy to parallelize parsers for large amounts of input data that is already given in a split format, e.g. a large list of individual database entries, or is easy to split by a fast preprocessing step, e.g. parsing the grammatical structure of sentences in large texts. A bit harder seems to be parallel parsing that already requires quite some effort to locate sub-structures in a given input. Common programming language code looks like a good example. In languages like Haskell, that

问题 It seems easy to parallelize parsers for large amounts of input data that is already given in a split format, e.g. a large list of individual database entries, or is easy to split by a fast preprocessing step, e.g. parsing the grammatical structure of sentences in large texts. A bit harder seems to be parallel parsing that already requires quite some effort to locate sub-structures in a given input. Common programming language code looks like a good example. In languages like Haskell, that

问题 I'm somewhat confused between the logic of calculating the height of binary tree. Code 1 public static int findHeight(Tree node) { if(node == null) return 0; else { return 1+Math.max(findHeight(node.left), findHeight(node.right)); } } Code 2 public static int findHeight(Tree node) { if(node == null) return -1; else { return 1+Math.max(findHeight(node.left), findHeight(node.right)); } } I think, the second one is correct, since it gives the correct answer for below code :- Tree t4 = new Tree(4

问题 How can i get the Parent in Binary tree in this code ? I wrote this : public class Node { public string state; public Node Left; public Node Right; public Node (string s , Node L , Node R ) { this.state = s; this.Right = R; this.Left = L; } public Node (string s) { this.state = s; this.Right = null; this.Left = null; } } And code this for tree of some data : 1 2 Now , how can i get the parent of Node like ( new Node("22lltrk", null, null) ) what i need to add to my code and where ? thanks .

问题 I'm learning about graphs(they seem super useful) and was wondering if I could get some advice on a possible way to structure my graphs. Simply, Lets say I get purchase order data everyday and some days its the same as the day before and on others its different. For example, yesterday I had an order of pencils and erasers, I create the two nodes to represent them and then today I get an order for an eraser and a marker, and so on. After each day, my program also looks to see who ordered what,

问题 I am trying to develop a solution to rotating a 2D array by 90 degrees, but the rotating to right or left would be dependent on the second argument of direction that is one of my challenges in addition to the fact that its rotating a 4 by 3 matrix. This is what I have thus far: const solve = (intArray, direction) => { const matrix = intArray.slice(); for (let rowIndex = 0; rowIndex < matrix.length; rowIndex+= 1) { for (let columnIndex = rowIndex + 1; columnIndex < matrix.length; columnIndex +

问题 I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it. Thanks of @LutzL's suggestion. I thought of stirling's approximation before. 回答1: The O -part should be easy. Choosing exactly n /2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not. The Ω -part is harder. In fact, plotting 4n / binomial(2 n

问题 Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). Question link is here The recursion method need to traverse the tree twice. But one of the comment provided a solution used a technique called 'Null check'. I can't understand why in this way can we avoid checking the tree twice? Here is his code in C++: bool isSymmetric(TreeNode* root) { if (!root) return true; return isSymmetric(root->left, root->right); } bool isSymmetric(TreeNode* t1, TreeNode*

问题 I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it. Thanks of @LutzL's suggestion. I thought of stirling's approximation before. 回答1: The O -part should be easy. Choosing exactly n /2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not. The Ω -part is harder. In fact, plotting 4n / binomial(2 n