工具类 AStarUtil.java import java.util.*; import java.util.stream.Collectors; /** * A星算法工具类 */ public class AStarUtil { private int[][] NODES; public AStarUtil(int[][] NODES) { this.NODES = NODES; } public AStarUtil() { } public static final int STEP =
Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 89317 Accepted Submission(s): 24279 Problem Description The doggie found a bone in an ancient maze, which fascinated him a
D - 最少花费 Time Limit: 30000/10000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others) Submit Status Bob是一个魔法师.这天,他被N座山所阻挡,这些山排成一排,每座山有一个高度.Bob需要从左往右每次跳到相邻的一座山上,相邻的两座山的高度差不能超过K,当从高度差为D的一座山跳到另一座山时要花A×D的魔法值.Bob可以改变一座山的高度,但调整后的高度不能小于0或大于1000,改变S
You are standing at position 0 on an infinite number line. There is a goal at position target. On each move, you can either go left or right. During the n-th move (starting from 1), you take n steps. Return the minimum number of steps required to rea
A move consists of taking a point (x, y) and transforming it to either (x, x+y) or (x+y, y). Given a starting point (sx, sy) and a target point (tx, ty), return True if and only if a sequence of moves exists to transform the point (sx, sy) to (tx, ty
Markdown版本笔记 我的GitHub首页 我的博客 我的微信 我的邮箱 MyAndroidBlogs baiqiantao baiqiantao bqt20094 baiqiantao@sina.com 领扣-754 到达终点数字 Reach a Number MD 目录 目录到达终点数字 Reach a Number MD题目思路代码实现 到达终点数字 Reach a Number MD 题目 在一根无限长的数轴上,你站在0的位置.终点在target的位置. You are standi
前两天面试遇到的一个题,当时没有想清楚,今天想了一下,po出来: # -*-encoding:utf-8-*- import sys end = 0 # 终点 cnt = 0 # 统计组合方式 def jump(start): global cnt for i in [1,2]: cur = str(start)+"+"+str(i) if eval(cur) >= end: print cur cnt += 1 continue jump(cur) def main(n): &
原题地址: https://oj.leetcode.com/problems/dungeon-game/ 题目内容: The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was init
1,每个小孩的糖果数量是多少 有p个小孩,c个糖果,刚开始第1个小孩发一个糖果,第2个小孩发两个糖果,第p个小孩发p个糖果,如果糖果没有发完,就接着[注意]第1个小孩发p+1个糖果.....第p个小孩发2*p个糖果,如果糖果还没有发完,就继续下一轮,第1个小孩发2*p+1个糖果,... 方法一,遍历[这种方法太low了就折叠一下] package com.cnblogs.mufasa.Main1; import java.util.Scanner; public class Main { pub