从Java中传递的字符串中获取最长的子字符串

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11 年前已关闭。

Answer 1

遵循下面提到的算法子字符串序列

Answer 2

我还没有测试过这个，我已经发现并修复了一个错误，但我想我现在已经涵盖了所有基础。您在问题中没有详细说明的一个限制是存在两个长度相等的子字符串的情况。即abbbcccdef，我只返回第一个。

public String getLongestSubstring(String s)
{
    if (s.length() == 0) return null;

    //We need some data to start with.
    char currentChar = s.charAt(0);
    String longestString = "" + currentChar;
    String runningString = "" + currentChar;
    int currentLongestLength = 1;

    for (int i = 1; i < s.length(); i++)
    {
        //Check the current char, is it the same?
        if (s.charAt(i) == currentChar)
        {
            runningString = runningString + currentChar;
            //Have we beaten our previous record.
            if (runningString.length() > longestString.length())
            {
                //Increase the record.
                currentLongestLength++;
                longestString = runningString;
            }
        }
        else
        {
            runningString = "" + s.charAt(i); 
        }
    }
    return longestString;
}

Answer 3

以下是我如何实现从字符串中查找最长的子字符串。

public class LongestString {    

    public static void main (String[] args) throws java.lang.Exception {
        System.out.println("Longest Substring is " + getLongestSubstring("abbcccdf"));
    }

    public static String getLongestSubstring(String s) {
            int length = 1;

        String longestString = "";

        for (int i = 0; i < s.length(); i++) {
            StringBuilder str = new StringBuilder();
            str.append(s.charAt(i));

            for (int j = i + 1; j < s.length(); j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    str.append(s.charAt(j));
                } else {
                    break;
                }
            }
            if (length < str.length()) {
                length = str.length();
                longestString = str.toString();
            }

        }
        return longestString;
    }
}

测试一下： http://ideone.com/JM904Y

Answer 4

public String getLongestSubString(String source) {

    StringBuilder longestStringBuilder = new StringBuilder(source.length());
    String longestString = "";

    for(int i=0; i<source.length(); i++) {

        if(longestStringBuilder.toString().contains(String.valueOf(source.charAt(i)))) {

            if(longestString.length() < longestStringBuilder.toString().length()) {
                longestString = longestStringBuilder.toString();
            }

            longestStringBuilder.setLength(0);
        }

        longestStringBuilder.append(source.charAt(i));

    }

    if(longestString.length() < longestStringBuilder.toString().length()) {
        longestString = longestStringBuilder.toString();
    }

    return longestString;
}