百度笔试题11.2

题目五：

http://topic.youkuaiyun.com/u/20070930/11/fa87b186-148e-4cc3-a582-013562e64de0.html

求给定的字符串中最长回文的长度。

例如给定的字符串为：“abcbaba”，最长的回文为“abcba”,长度为5

 

Answer1:

O(nlogn)的算法参见 参考资料里面的《后缀数组+最长回文之串.pdf》

字符串s1的反转串s2,求s1和s2的最长公共子串。利用求最长公共子串的方法，还是采用后缀数组的数据结构，时间复杂度为O(nlogn)

 

 

Answer2:

O(nlogn)的算法

http://hi.baidu.com/cuifenghui/blog/item/a665553d10720ded3c6d97ec.html

/*

思路描述: 最长回文串是关于中心轴对称的，可以分成求

最长奇回文子串和最长偶回文子串。对于最长奇回文子串

可以从中间一个点向两边延伸的方法来找出以该点为中心

的最长奇回文串，通过枚举中心点的方法求出整个串最长

的奇回文串；对于偶回文串，可以通过给原字符串加入一些

"空白字符"组成一个新字符串，然后利用求奇回文串的方法

做出。本程序就是利用填充"空白字符"并枚举中心点的方法

来找出字符串中最长的回文子串，时间复杂度为o(n^2)。

*/

 

Answer3:

http://hi.baidu.com/guidee/blog/item/c8ce40cba18eba17be09e6b1.html

/*

         * 题目：

         * 1：判断字符串是否是回文

         * 2：字符串最长回文子串

         *

         * */

 

        public class Symmetric

        {

            public Symmetric()

            {

            }

            /// <summary>

            /// 1：判断字符串是否是回文

            /// </summary>

            /// <param name="str"></param>

            /// <returns></returns>

            public bool IsSymmetric(string str)

            {

                int i = 0;

                int j = str.Length - 1;

                while (i < j)

                {

                    if (str[i] != str[j])

                        return false;

                    i++;

                    j--;

                }

                return true;

            }

            /// <summary>

            /// 2：字符串最长回文子串，方法一：从中心开始向两边搜索对称字符。较高效

            /// </summary>

            /// <param name="str"></param>

            /// <param name="startPos"></param>

            /// <returns></returns>

            public int GetMaxSymmetricLengthAndPos1(string str, out int startPos)

            {

                startPos = -1;

                if (string.IsNullOrEmpty(str))

                {

 

                    return 0;

                }

 

                int sameStartPos = 0;

                int sameEndPos = 0;

                int maxLen = 0;

                while (sameStartPos < str.Length)

                {

                    sameEndPos = sameStartPos;

                    //判断有没有连续相同的字符，没有，则直接跳过

                    while (sameEndPos + 1 < str.Length && str[sameEndPos + 1] == str[sameStartPos])

                        sameEndPos++;

                    //寻找重复字符两边对称的字母

                    int searchLeft = sameStartPos;

                    int searchRight = sameEndPos;

                    while (searchLeft - 1 >= 0 && searchRight + 1 < str.Length

                        && str[searchLeft - 1] == str[searchRight + 1])

                    {

                        searchLeft--;

                        searchRight++;

                    }

                    //判断是否长度更大

                    if (searchRight - searchLeft + 1 > maxLen)

                    {

                        //保存更大长度和对应的起始位置

                        maxLen = searchRight - searchLeft + 1;

                        startPos = searchLeft;

                    }

                    //设置搜做新位置，重复字符后面第一个位置

                    sameStartPos = sameEndPos + 1;

                }

                return maxLen;

            }

 

            /// <summary>

            /// 2：字符串最长回文子串，方法二：从两边开始向中心搜索对称字符。低效

            /// </summary>

            /// <param name="str"></param>

            /// <param name="startPos"></param>

            /// <returns></returns>

            public void GetMaxSymmetricLengthAndPos2(string str, out int sameStartPos, out int count)

            {

                int begin = 0;

                int end = str.Length - 1;

                int max = 0;

                count = 0;

                sameStartPos = -1;

 

                int max = 0;

 

                for (; begin < str.Length - 1; begin++)

                {

                    for (; end > begin; end--)

                    {

                        if (IsSymmetric(str, begin, end))

                        {

                            max = end - begin;

                            if (max > count)

                            {

                                count = max;

                                sameStartPos = begin;

 

                            }

                            break;

                        }

                    }

                }

            }

 

            /// <summary>

            /// 判断子串是否回文

            /// </summary>

            /// <param name="str"></param>

            /// <param name="begin"></param>

            /// <param name="end"></param>

            /// <returns></returns>

            public bool IsSymmetric(string str, int begin, int end)

            {

                int i = begin, j = end;

                for (; i < j; i++, j--)

                {

                    if (str[i] != str[j])

                    {

                        return false;

                    }

                }

                return true;

            }

 

        }

}