最长运行C#的索引
本文关键字:索引 运行 | 更新日期: 2023-09-27 18:21:43
我正在努力解决这个问题:编写一个函数,查找字符串中最长游程的从零开始的索引。跑步是同一个字符的连续序列。如果有多个具有相同长度的游程,则返回第一个游程的索引。
例如,IndexOfLongestRun("abbcccdddcccbba")应该返回6,因为最长的运行是dddd,并且它首次出现在索引6上。
以下是我所做的:
private static int IndexOfLongestRun(string str)
{
char[] array1 = str.ToCharArray();
//Array.Sort(array1);
Comparer comparer = new Comparer();
int counter =1;
int maxCount = 0;
int idenxOf = 0;
for (int i =0; i<array1.Length-1 ; i++)
{
if (comparer.Compare(array1[i],array1[i+1]) == 0)
{
counter++;
}
else {
if(maxCount < counter)
{
maxCount = counter;
idenxOf = i - counter + 1;
}
counter = 1;
}
}
return idenxOf ;
}
}
public class Comparer : IComparer<char>
{
public int Compare(char firstChar, char nextChar)
{
return firstChar.CompareTo(nextChar);
}
}
问题是,当我到达最后一个索引时,例如"abbccaaaaaaaa"在这种情况下,它是一个,当i=14
(以这个字符串为例)和i<array1.Length-1
语句为假时,for循环直接跳到return indexOf;
并返回错误的索引,我正试图找出如何推送forloop来继续实现,以便将idenxOf更改为正确的索引。请帮忙吗?
当current==previor时,您可以检查每个迭代是否获得了新的最佳分数。速度最低,但它允许您通过省略循环后的额外检查来编写较短的代码:
int IndexOfLongestRun(string input)
{
int bestIndex = 0, bestScore = 0, currIndex = 0;
for (var i = 0; i < input.Length; ++i)
{
if (input[i] == input[currIndex])
{
if (bestScore < i - currIndex)
{
bestIndex = currIndex;
bestScore = i - currIndex;
}
}
else
{
currIndex = i;
}
}
return bestIndex;
}
将循环变量i提升到方法范围,并在循环退出后立即重复条件块if(maxCount<counter){…}。因此,它在循环完成后再执行一次
private static int IndexOfLongestRun(string str)
{
char[] array1 = str.ToCharArray();
//Array.Sort(array1);
Comparer comparer = new Comparer();
int counter = 1;
int maxCount = 0;
int idenxOf = 0;
int i;
for (i = 0; i < array1.Length - 1; i++)
{
if (comparer.Compare(array1[i], array1[i + 1]) == 0)
{
counter++;
}
else
{
if (maxCount < counter)
{
maxCount = counter;
idenxOf = i - counter + 1;
}
counter = 1;
}
}
if (maxCount < counter)
{
maxCount = counter;
idenxOf = i - counter + 1;
}
return idenxOf;
}
像往常一样迟到了,但加入了聚会。一种自然的经典算法:
static int IndexOfLongestRun(string input)
{
int longestRunStart = -1, longestRunLength = 0;
for (int i = 0; i < input.Length; )
{
var runValue = input[i];
int runStart = i;
while (++i < input.Length && input[i] == runValue) { }
int runLength = i - runStart;
if (longestRunLength < runLength)
{
longestRunStart = runStart;
longestRunLength = runLength;
}
}
return longestRunStart;
}
最后,你有最长的跑步指数和长度。
public static int IndexOfLongestRun(string str)
{
var longestRunCount = 1;
var longestRunIndex = 0;
var isNew = false;
var dic = new Dictionary<int, int>();
for (var i = 0; i < str.Length - 1; i++)
{
if (str[i] == str[i + 1])
{
if (isNew) longestRunIndex = i;
longestRunCount++;
isNew = false;
}
else
{
isNew = true;
dic.Add(longestRunIndex, longestRunCount);
longestRunIndex = 0;
longestRunCount = 1;
}
}
return dic.OrderByDescending(x => x.Value).First().Key;
}
如果字符串为空,则返回-1,并且您可以根据规范灵活地返回索引和计数。
string myStr = "aaaabbbbccccccccccccdeeeeeeeee";
var longestIndexStart = -1;
var longestCount = 0;
var currentCount = 1;
var currentIndexStart = 0;
for (var idx = 1; idx < myStr.Length; idx++)
{
if (myStr[idx] == myStr[currentIndexStart])
currentCount++;
else
{
if (currentCount > longestCount)
{
longestIndexStart = currentIndexStart;
longestCount = currentCount;
}
currentIndexStart = idx;
currentCount = 1;
}
}
return longestIndexStart;
Kvam接受的答案非常适合小字符串,但随着长度接近100000个字符(也许不需要),其效率会下降。
public static int IndexOfLongestRun(string str)
{
Dictionary<string, int> letterCount = new Dictionary<string, int>();
for (int i = 0; i < str.Length; i++)
{
string c = str.Substring(i, 1);
if (letterCount.ContainsKey(c))
letterCount[c]++;
else
letterCount.Add(c, 1);
}
return letterCount.Values.Max();
}
这种解决方案的速度是Kvam的两倍。也许还有其他优化。