Regionalliga Southwest stats & predictions

Understanding Betting Odds

Betting odds provide insights into how bookmakers perceive each team's chances of winning. Understanding these odds can help you make more informed betting decisions. For instance, if a team is listed as an underdog but you believe they have a strong chance due to recent form or home advantage, you might consider placing a bet on them.

Betting on Draws

In leagues like the Regionalliga Southwest Germany, where matches are often tightly contested, betting on draws can be a viable strategy. Teams are generally well-matched in terms of skill level, leading to many games ending in draws or low-scoring results.

Betting on Total Goalsmengyuzhi/algorithm-study<|file_sep|>/algorithm-study/src/main/java/com/leisure/study/algorithm/offer/offer_18.java package com.leisure.study.algorithm.offer; /** * Created by mengyuzhi on Aug/19/2017. * 题目描述 * 已知一个字符串的编码为： * "A" -> "1" * "B" -> "2" * ... * "Z" -> "26" * 给定一个数字字符串，按上述规则反向解码，求出所有解码的方法数。 */ public class offer_18 { public int getNum(String str) { if(str == null || str.length() ==0) return -1; int[] dp = new int[str.length()+1]; dp[0] = dp[1] =1; for (int i =2;i<=str.length();i++){ if(str.charAt(i-1) != '0'){ dp[i] +=dp[i-1]; } if(str.charAt(i-2) == '1' ||(str.charAt(i-2) == '2' && str.charAt(i-1)<='6')){ dp[i] +=dp[i-2]; } } return dp[str.length()]; } } <|repo_name|>mengyuzhi/algorithm-study<|file_sep|>/algorithm-study/src/main/java/com/leisure/study/algorithm/sort/bubble_sort.java package com.leisure.study.algorithm.sort; import java.util.Arrays; /** * Created by mengyuzhi on Aug/25/2017. * 冒泡排序算法 */ public class bubble_sort { public static void main(String[] args){ int[] arr = {5 ,6 ,8 ,9 ,3 ,2 ,7}; bubble_sort(arr); System.out.println(Arrays.toString(arr)); } private static void bubble_sort(int[] arr){ boolean flag = true; for(int i=0;iarr[j+1]){ int tmp = arr[j]; arr[j] = arr[j+1]; arr[j+1] = tmp; flag = false; } } if(flag) break; } } } <|repo_name|>mengyuzhi/algorithm-study<|file_sep|>/algorithm-study/src/main/java/com/leisure/study/algorithm/dynamic_programming/CoinChange.java package com.leisure.study.algorithm.dynamic_programming; /** * Created by mengyuzhi on Aug/19/2017. * * 给定不同面额的硬币 coins 和一个总金额 amount。 * 编写一个函数来计算可以凑成总金额所需的最少的硬币个数。 如果没有任何一种硬币组合能组成总金额，返回 -1。 * * Example: * * coins = [1,2,5], amount =11 * * return 3 （11 = 5 + 5 + 1） */ public class CoinChange { public int coinChange(int[] coins,int amount){ if(coins==null || coins.length ==0) return -1; int[] dp = new int[amount+1]; dp[0] =0; for (int i=1;i<=amount;i++){ int minNum = Integer.MAX_VALUE; for (int j=0;j=0 && dp[i-coins[j]] != Integer.MAX_VALUE){ minNum = Math.min(minNum ,dp[i-coins[j]]+1); } } dp[i] = minNum; } return dp[amount]==Integer.MAX_VALUE?-1:dp[amount]; } } <|repo_name|>mengyuzhi/algorithm-study<|file_sep|>/algorithm-study/src/main/java/com/leisure/study/algorithm/dynamic_programming/LadderLength.java package com.leisure.study.algorithm.dynamic_programming; import java.util.ArrayList; import java.util.List; /** * Created by mengyuzhi on Aug/23/2017. * * 给定两个单词（beginWord 和 endWord）和一个字典 wordList， * 找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则： * * 每次转换只能改变一个字母。 * 转换过程中的中间单词必须是字典中的单词。 * * * * * 示例: * * 输入: * beginWord = "hit", * endWord = "cog", * wordList = ["hot","dot","dog","lot","log","cog"] * * 输出: 5 * * 解释: 一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", * 返回它的长度 5。 * */ public class LadderLength { private List[] buildMap(String[] words){ List[] map = new ArrayList[words.length]; for (String word : words){ map[word.hashCode()] = new ArrayList<>(); } for (int i=0;i[] map = buildMap(wordList); // System.out.println(map[beginWord.hashCode()]); // System.out.println(map[endWord.hashCode()]); // return bfs(map,endWord,new HashSet<>(Arrays.asList(beginWord))); // return dfs(map,endWord,new HashSet<>(Arrays.asList(beginWord)),0); return bfs2(map,endWord,new HashSet<>(Arrays.asList(beginWord))); } private int bfs(List[] map,String endWord,List visited){ List queue = new ArrayList<>(); queue.add(new String[]{endWord}); int level=0; while(!queue.isEmpty()){ level++; List nextQueue=new ArrayList<>(); for (String[] strs : queue){ String lastStr=strs[strs.length-1]; List nextStrs=map[lastStr.hashCode()]; for (String str : nextStrs){ if(!visited.contains(str)){ visited.add(str); String[] nextArr=new String[strs.length+1]; System.arraycopy(strs,nextArr,strs.length-nextArr.length,strs.length-nextArr.length,strs.length); nextArr[strs.length]=str; if(str.equals(endWord)) return level+1; nextQueue.add(nextArr); } } } queue=nextQueue; } return -1; } private int dfs(List[] map,String endWord,List visited,int level){ List nextStrs=map[endWord.hashCode()]; if(nextStrs==null || nextStrs.size()==0) return -1; level++; for (String str : nextStrs){ if(!visited.contains(str)){ visited.add(str); if(str.equals(endWord)) return level+1; int ret=dfs(map,endWord,new HashSet<>(visited),level); if(ret!=-1) return ret; } } return -1; } private int bfs2(List[] map,String endWord,List visited){ List queue=new ArrayList<>(); queue.add(new String[]{endWord}); int level=0; while(!queue.isEmpty()){ level++; List nextQueue=new ArrayList<>(); for (String[] strs : queue){ String lastStr=strs[strs.length-1]; List nextStrList=map[lastStr.hashCode()]; for (String str : nextStrList){ if(!visited.contains(str)){ visited.add(str); String[] nextArr=new String[strs.length+1]; System.arraycopy(strs,nextArr,strs.length-nextArr.length,strs.length-nextArr.length,strs.length); nextArr[strs.length]=str; if(str.equals(endWord)) return level+1; nextQueue.add(nextArr); } } } queue=nextQueue; } return -1; } } <|repo_name|>mengyuzhi/algorithm-study<|file_sep|>/algorithm-study/src/main/java/com/leisure/study/algorithm/dynamic_programming/NthUglyNumber.java package com.leisure.study.algorithm.dynamic_programming; /** * Created by mengyuzhi on Aug/24/2017. */ public class NthUglyNumber { // public static void main(String[] args) { // NthUglyNumber nuthUglyNumber=new NthUglyNumber(); // System.out.println(nuthUglyNumber.nthUglyNumber(10)); // } public int nthUglyNumber(int n) { long [] ugly=new long[n]; // ugly数组保存前n个丑数，ugly[0]=第一个丑数为1 ugly[0]=1; // 设置初始值 // 记录ugly数组下标 int index_2=0; //记录以2为因子生成丑数时的下标，初始值为第一个丑数下标（即为0） int index_3=0; //记录以3为因子生成丑数时的下标，初始值为第一个丑数下标（即为0） int index_5=0; //记录以5为因子生成丑数时的下标，初始值为第一个丑数下标（即为0） //从第二个开始到第n个 for(int i=1;i