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My solution for hackerrank puzzles. Contribute to santik/hackerrank development by creating an account on GitHub.
queryURL = removeSlash(queryURL);//Check Fieldsif(fields.size() >0) { queryURL = queryURL.concat("?fields="+StringUtils.join(fields,",") +"&"+ WoWAPI.get().getURLSuffix().substring(1)); }else{ queryURL = queryURL.concat(WoWAPI.get().getURLSuffix()); } } } 开发者ID:M-AJ...
publicStringsubstring(intbeginIndex);publicStringsubstring(intbeginIndex,intendIndex); 第二十三关:Java Substring Comparisons 连续字符字典序排序,输出最大和最小值 Sample Input 0 welcometojava 3 Sample Output 0 ava wel 解决方案: importjava.util.Scanner;publicclassSolution{publicstaticStringgetSmallestAndLarges...
开发者ID:rshaghoulian,项目名称:HackerRank_solutions,代码行数:54,代码来源:Solution.java 示例12: computeNeg ▲点赞 2▼ importjava.util.BitSet;//导入方法依赖的package包/类/** Returns the negation of a (bit) set. */privateBitSetcomputeNeg(BitSet arg){ ...
bMap.put(b.substring(i,i+1), bar); } java.util.HashSet<String> aSet = new java.util.HashSet<String>(aMap.keySet()); java.util.HashSet<String> bSet = new java.util.HashSet<String>(bMap.keySet()); if (aSet.containsAll(bSet) && bSet.containsAll(aSet)) { ...
Hackerrank Practice Anagram 拆分数组 看一半操作几次能成为另一半的anagram 题目 输入第一行是要判断的字符串的个数n,之后的n行输入为需要判断的字符串。 每个字符串str,是两个等长字符串的合体,所以长度一定是偶数。若为奇数,返回-1。 所以str分成两个substring,右边需要多少次操作能变为左边。只要用一个26位...
currInr = "Rs." + currInr.substring(1); } String currUs = usdFormatter.format(amt); String currFr = eurFormatter.format(amt); String currCn = cnyFormatter.format(amt); System.out.println("US: " + currUs); System.out.println("India: " + currInr); ...
To ensure this time complexity, utilizing a balanced BST such as a Red Black Tree is necessary. However, it's important to keep in mind that online judges like Codeforce or Hackerrank often furnish inputs that result in degenerate BSTs. ...
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