LeetCode0005.最长回文子串

题目要求

算法分析

暴力算法

截取每一个子串 判断是否为回文串 输出其中最长的回文子串。

中心扩展

若回文串两侧的字符相同，则删去两侧字符后剩下的字符串仍为回文串。

我们可以反过来，对字符串中的某一位字符c(或某两位相邻的相同字符串cc)，

如果其两侧字符相同，则让它向两侧扩展，直到两侧字符不同，最后便获得了以c(或cc)为中心的回文字串。

对每一位都进行上述操作,便可获取最长的回文子串。

代码展示(C#)

public string LongestPalindrome(string s)
        {
            if(s.Length < 2)
            {
                return s;
            }
            string result = "";
            int maxLength = 0;
            //"abcdefg"
            // i
            for (int i = 0; i < s.Length; i++) {
                for (int j = 1; j <= s.Length - i; j++)
                {
                    string temp = s.Substring(i, j);
                    if (IsPalindrome(temp)&& j > maxLength)
                    {
                        result = temp;
                        maxLength = j;
                    }
                }
            }

            return result;
        }

        public bool IsPalindrome(string s)
        {
            bool result = true;
            int i = 0;
            int j = s.Length - 1;
            while (i < j)
            {
                if (s[i++] != s[j--])
                {
                    result = false;
                    break;
                }
            }
            return result;
        }

public class Solution {
    public string LongestPalindrome(string s)
    {
        if (s.Length < 2)
        {
            return s;
        }
        string result = "";
        int max = 0;
        int start = 0;
        //cbbd
        for (int i = 0; i < s.Length; i++)
        {
            int length1 = OneCentralExpansion(s, i);
            if (length1 > max)
            {
                max = length1;
                start = i - (max - 1) / 2;
            }
            if (i < s.Length - 1 && s[i] == s[i+1])
            {
                int length2 = TwoCentralExpansion(s, i);

                if (length2 > max)
                {
                    max = length2;
                    start = i - (max - 1) / 2;
                }
            }
        }
        result = s.Substring(start, max);
        return result;
    }

    /// <summary>
    /// 单字符扩展
    /// </summary>
    /// <param name="s">原字符串</param>
    /// <param name="center">子串中心字符的索引</param>
    /// <returns>回文子串长度</returns>
    public int OneCentralExpansion(string s,int center)
    {
        int result = 1;
        int left = center - 1;
        int right = center + 1;
        while (left >= 0 && right < s.Length)
        {
            if (s[left--] == s[right++])
            {
                result += 2;
            }
            else
            {
                break;
            }
        }
        return result;
    }

    /// <summary>
    /// 双字符扩展
    /// </summary>
    /// <param name="s">原字符串</param>
    /// <param name="center">子串中心字符串的第一个字符的索引</param>
    /// <returns>回文子串长度</returns>
    public int TwoCentralExpansion(string s, int center)
    {
        int result = 1 +1;
        int left = center - 1;
        int right = center + 1  +1;
        while (left >= 0 && right < s.Length)
        {
            if (s[left--] == s[right++])
            {
                result += 2;
            }
            else
            {
                break;
            }
        }
        return result;
    }
}

代码展示(C++)

class Solution {
public:
    string longestPalindrome(string s) {
        if (s.size() < 2) {
            return s;
        }
        string result = "";
        int max = 0;
        int start = 0;
        for (int i = 0; i < s.size(); ++i) {
            int length1 = CentralExpansion(s, i, 0);
            if (length1 > max) {
                max = length1;
                start = i - (max - 1) / 2;
            }
            if (i < s.size() - 1 && s[i] == s[i + 1]) {
                int length2 = CentralExpansion(s, i, 1);
                if (length2 > max) {
                    max = length2;
                    start = i - (max - 1) / 2;
                }
            }
        }
        result = s.substr(start, max);
        return result;
    }

    int CentralExpansion(string s, int center, int plus) {
        int result = 1 + plus;
        int left = center - 1;
        int right = center + 1 + plus;
        while (left >= 0 && right < s.size()) {
            if (s[left--] == s[right++]) {
                result += 2;
            }
            else {
                break;
            }
        }
        return result;
    }
};

提交结果

 

 

一些感想

代码写的过于冗杂,需要优化