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38. Search a 2D Matrix II https://www.lintcode.com/problem/search-a-2d-matrix-ii/description?_from=ladder&&fromId=1 这道题与二分法有什么关系呢? -把整个二维数组从对角线分成两半,从左下角开始,往右上角逼近. public class Solution { /** * @param matrix: A list of lists of integers * @param ta…
254. Drop Eggs https://www.lintcode.com/problem/drop-eggs/description?_from=ladder&&fromId=1 28. Search a 2D Matrix https://www.lintcode.com/problem/search-a-2d-matrix/description?_from=ladder&&fromId=1 思路1: 1. find the row index, the last…
1.二叉搜索树:去一个陌生的城市问路到目的地: for each node, all elements in its left subtree are less-or-equal to the node (<=), and all the elements in its right subtree are greater than the node (>). 给予一个已经排序好的整数数组, 生成一个相对合理的二叉搜索树.(?相对合理的?) 给予一个二叉树的根节点,验证该树是否是二叉树搜索树,(…
(binary search trees) which form the basis of modern databases and immutable data structures. Binary search works very much the way humans intuitively search for a name in a yellow pages directory (if you have ever seen one) or the dictionary. In thi…
Complexity: O(log(n)) Ref: Binary search algorithm or 二分搜索算法 Ref: C 版本 while 循环 C Language scripts by McDelfino: #include <stdio.h> #include <stdlib.h> int getIndex(int* arr, int length, int num); int main() { int arr[100]; for (int i = 0; i &…
js binary search algorithm js 二分查找算法 二分查找, 前置条件 存储在数组中 有序排列 理想条件: 数组是递增排列,数组中的元素互不相同; 重排 & 去重 顺序: 递增排列/递减排列; 重复: 数组中存在相同的元素; "use strict"; /** * * @author xgqfrms * @license MIT * @copyright xgqfrms * @created 2020-05-20 * @modified * * @des…
The solution for the problem can be divided into three cases: case 1: if the delete node is leaf node, then we can simply remove it case 2: if the delete node is has single side or substree case 3: it has two children, then to keep it as a valid bina…
What is Binary Search Tree (BST) A binary tree in which for each node, value of all the nodes in left subtree is less or equal and value of all the nodes in right subtree is greater The idea: We can use set boundry for each node. We take C tree for e…
Let's say we are going to find out number of occurrences of a number in a sorted array using binary search in O(log n) time. For example the given array is: [1,1,3,5,5,5,5,5,9,11], the number 5 appears 5 times; the number 3 appears 1 time; 2 appears…
If you understand the comments below, never will you make mistakes with binary search! thanks to A simple CPP solution with lower_bound and C++ O(logn) Binary Search that handles duplicate, thanks to phu1ku 's answer on the second post. http://en.cpp…