编码挑战：安排数字以形成最大可能的整数

本文介绍了编码挑战：安排数字以形成最大可能的整数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 今天的后期编码挑战很简单。 给定一组整数，排列整数，以便这样形成的最终整数是可能的最大整数。 例如。 {1,5,67,9}将被安排成96751 该列表最多可包含50个数字，每个数字将小于256且为正数。 我尝试过： 为了保持问题的含糊不清。 上周的获胜者是Graeme_Grant，主要是因为他教给我们所有我们从未知道的单词。向Sean发送你的详细信息和适当的东西（可能）将会回家。The late coding challenge for today is straightforward.Given a set of integers, arrange the integers so that the final integer thus formed is the largest number possible.For example.{ 1,5,67,9 } would be arranged to form 96751The list may contain up to 50 numbers, and each number will be less than 256 and positive.What I have tried:To keep the problems suitably ambiguous.Last week's winner was Graeme_Grant, mainly because he taught us all words we never knew. Send Sean your details and something (possibly) appropriate will wing it's way home.推荐答案 System.Console.WriteLine ( BiggestIntegerInator( System.Environment.CommandLine.Rive ( Option.RemoveEmptyEntries | Option.RemoveQuotes | Option.HonorEscapes | Option.HonorQuotes )) ) ; （Rive在我的一篇文章中，它是Split的一个更灵活的版本。） 1 05 6.7E + 19收益率 96751 (Rive is in one of my Articles, it's a more flexible version of Split.)1 05 6.7E+1 "9" yields 96751private static stringBiggestIntegerInator( System.Collections.Generic.IList<string> Values){ System.Text.StringBuilder result = new System.Text.StringBuilder() ; System.Collections.Generic.SortedList<string,int> l = new System.Collections.Generic.SortedList<string,int> ( Values.Count , new DescendingComparer<string>() ) ; for ( int i = 0 ; i < Values.Count ; i++ ) { string v = Values [ i ] ; double j ; if ( System.Double.TryParse ( v , out j ) && ( ( v = j.ToString() ).IndexOf ( '.' ) == -1 ) ) { if ( l.ContainsKey ( v ) ) { l [ v ]++ ; } else { l [ v ] = 1 ; } } } for ( int i = 0 ; i < l.Count ; i++ ) { string v = l.Keys [ i ] ; for ( int j = 0 ; j < l [ v ] ; j++ ) { result.Append ( v ) ; } } return ( result.ToString() );}private class DescendingComparer<T> : System.Collections.Generic.IComparer<T>where T : System.IComparable<T>{ public int Compare ( T Op0 , T Op1 ) { return ( Op1.CompareTo ( Op0 ) ) ; }} 好​​的，所以上面的比较器不太合适，这是另一个： OK, so that comparer above doesn't quite do the trick, here's another:private class DescendingNumericStringComparer : System.Collections.Generic.IComparer<string>{ public int Compare ( string Op0 , string Op1 ) { int result = 0 ; int i0 = -1 ; int i1 = -1 ; while ( ( result == 0 ) && ( ( i0 < Op0.Length - 1 ) || ( i1 < Op1.Length - 1 ) ) ) { if ( i0 < Op0.Length - 1 ) i0++ ; if ( i1 < Op1.Length - 1 ) i1++ ; result = Op1 [ i1 ].CompareTo ( Op0 [ i0 ] ) ; } if ( result == 0 ) { result = Op0.Length.CompareTo ( Op1.Length ) ; } return ( result ) ; }} 43 432 435 433 收益 435 43 433 432 所以也是一个缺陷。这是一个肮脏的小修复（不建议，由于字符串操作），而我正在进行更好的实现： 43 432 435 433 yields 435 43 433 432So that one has a flaw as well. Here's a dirty little fix (not recommended, due to string manipulation) while I work on a better implementation:private class DescendingStringComparer : System.Collections.Generic.IComparer<string>{ public int Compare ( string Op0 , string Op1 ) { int result = (Op1 + Op0).CompareTo ( Op0 + Op1 ) ; if ( result == 0 ) { result = Op0.Length.CompareTo ( Op1.Length ) ; } return ( result ) ; }} 24 242 243 收益 243 24 242 第四个比较器，没有字符串连接。 获取两个长度中的较长者并迭代。 当我们用其中一个值中的字符用完时，开始使用另一个值开头的字符。 24 242 243 yields 243 24 242Fourth comparer, no string concatenation.Get the longer of the two lengths and iterate.When we run out of characters in one of the values, start using the characters at the start of the other value instead.private class DescendingStringComparer : System.Collections.Generic.IComparer<string>{ public unsafe int Compare ( string Op0 , string Op1 ) { int result = 0 ; int len = Op0.Length > Op1.Length ? Op0.Length : Op1.Length ; for ( int i = 0 ; ( result == 0 ) && ( i < len ) ; i++ ) { char c0 = i < Op0.Length ? Op0 [ i ] : Op1 [ i - Op0.Length ] ; char c1 = i < Op1.Length ? Op1 [ i ] : Op0 [ i - Op1.Length ] ; result = c1.CompareTo ( c0 ) ; } if ( result == 0 ) { result = Op0.Length.CompareTo ( Op1.Length ) ; } return ( result ) ; }} * CCCP Code Challenge Code Project* Build largest integerclearlargest({1, 5, 67, 9})largest({100, 11, 10, 110, 112,1,114})largest({4, 45, 46, 43})largest({43, 432, 435, 433})largest({2, 243, 242, 245, 241, 24, 221})largest({20, 221, 226, 202, 2, 201})procedure largest(lst)*convert 2 stringfor scan=1 to len(lst)lst[scan]= str(lst[scan],,,.T.)next?? lst[1]for scan=2 to len(lst)?? " "?? lst[scan]next*tri par insertionfor scan=2 to len(lst)for ptr= scan to 2 step -1*pad until same lentgh mx= max(len(lst[ptr]), len(lst[ptr-1]))+1tmp1= lst[ptr]while len(tmp1) < mxtmp1 += lst[ptr]enddotmp1= left(tmp1,mx)tmp2= lst[ptr-1]while len(tmp2) < mxtmp2 += lst[ptr-1]enddotmp2= left(tmp2,mx)if tmp1 > tmp2tmp= lst[ptr]lst[ptr]= lst[ptr-1]lst[ptr-1]= tmpendifnextnext*result? lst[1]for scan=2 to len(lst)?? " "?? lst[scan]nextreturn 没有填充的变体字符串： Variant without padding the strings:for scan=2 to len(lst) for ptr= scan to 2 step -1 if len(lst[ptr])= len(lst[ptr-1]) tmp1= lst[ptr] tmp2= lst[ptr-1] else mx= len(lst[ptr])+ len(lst[ptr-1])- 1 for scanl= 0 to mx tmp1= lst[ptr, scanl%len(lst[ptr])+1] tmp2= lst[ptr-1, scanl%len(lst[ptr-1])+1] if tmp1 != tmp2 exit endif next endif if tmp1 > tmp2 tmp= lst[ptr] lst[ptr]= lst[ptr-1] lst[ptr-1]= tmp endif nextnext 测试集 Test sets1 5 67 99 67 5 1100 11 10 110 112 1 114114 112 11 1 110 10 1004 45 46 4346 45 4 4343 432 435 433435 43 433 4322 243 242 245 241 24 221245 243 24 242 241 2 22120 221 226 202 2 201226 2 221 202 20 201 注意：答案中的空格只是帮助读出如何订购数字。 [更新]精炼排序部分 填充部分非常棘手，我到目前为止最好的猜测： 填充是直到最大长度+ 1 填充是通过重复该值来进行的。 取20和202.什么是Ť他订购了吗？ 填充：20 - > 2020,202 - > 2022 订单与2020年和2022年相同，所以202,20。 取24和242.订单是什么？ 填充：24 - > 2424,242 - > 2422 订单与2424和2422相同，所以24,242。Note: Spaces in answers just help reading out how numbers are ordered.[Update] Refined the sort partThe padding part is really tricky, my best guess so far:The padding is to be made until maximum length + 1The padding is to be made by repeating the value.Take 20 and 202. What is the order ?padding: 20 -> 2020, 202 -> 2022The order is same as for 2020 and 2022, so 202, 20.Take 24 and 242. What is the order ?padding: 24 -> 2424, 242 -> 2422The order is same as for 2424 and 2422, so 24, 242. public static BigInteger DoChallenge(this IEnumerable<byte> question){ return (from x in question orderby (x < 10) ? x * 1110 : (x < 100) ? x * 100 + Math.Min(x, (x % 10) * 10 + x / 10) : x * 10 descending select new BigInteger(x)) .Aggregate((x, y) => (y < 10) ? x * 10 + y : (y < 100) ? x * 100 + y : x * 1000 + y);}数学思想是将每个1-2位和3位数字转换为4位数字，以便进行排序： 123转到1230 12转到1212，在122（1220）和121（1210）之间切换 21转到2112，在212（2120）之间切换）和211（2110） 2进入2220，在223（2230）和221（2210）之间以及23（2323）和21（2112）之间插入 The math idea is to convert each 1- 2- and 3-digit number to a 4-digit number for the purposes of sorting:123 goes to 123012 goes to 1212, slotting it between 122 (1220) and 121 (1210)21 goes to 2112, slotting it between 212 (2120) and 211 (2110)2 goes to 2220, slotting it between 223 (2230) and 221 (2210), and between 23 (2323) and 21 (2112) 这篇关于编码挑战：安排数字以形成最大可能的整数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！