Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be us
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size). Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rul
转自licoolxue https://blog.csdn.net/licoolxue/article/details/1533364 int -> String int i=12345;String s="";第一种方法:s=i+""; 第二种方法:s=String.valueOf(i);这两种方法有什么区别呢?作用是不是一样的呢?是不是在任何下都能互换呢? String -> int s="12345";int i;第一种方法:i
偶然在群里看到有人问到大数据查询,自己也就想了小艾改如何解决,从从1000万个随机数中查找出相同的10万个随机数花的最少时间, 谈到效率,自然是hashmap莫属. import java.util.HashMap;import java.util.Map;import java.util.Random; public class dataTest { private static final int [] datas = new int [1000 * 10000]; //1000万个随机数
Given a list of directory info including directory path, and all the files with contents in this directory, you need to find out all the groups of duplicate files in the file system in terms of their paths. A group of duplicate files consists of at l