Leetcode: Longest Palindromic Substring

问题描述：
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: “babad”

Output: “bab”

Note: “aba” is also a valid answer.
Example:

Input: “cbbd”

Output: “bb”

解决方案：
求最大回文子串，我之前正好看过Manacher算法解决最大回文子串的问题，算法的时间复杂度仅为O(n)，这个算法可以将长度为奇数和偶数的回文子串统一处理，具体算法的描述请看我的文章http://blog.youkuaiyun.com/u013362955/article/details/71516107

代码

 string longestPalindrome(string str) {

    string str0;
    str0 += "$#";
    for(int i = 0; i < str.size(); i++)
    {
        str0 += str[i];
        str0 += "#";
    }

    string s = findBMstr(str0);
    return s;
    }


    string findBMstr(string& str)
{
    int *p = new int[str.size() + 1];
    memset(p, 0, sizeof(p));

    int mx = 0, id = 0;   //mx记录最大边界，id记录回文中心点
    for(int i = 1; i <=  str.size(); i++)
    {
        if(mx > i)   
        {
            p[i] = (p[2*id - i] < (mx - i) ? p[2*id - i] : (mx - i));
        }
        else
        {
            p[i] = 1;
        }

        while(str[i - p[i]] == str[i + p[i]])
            p[i]++;

        if(i + p[i] > mx)
        {
            mx = i + p[i];
            id = i;
        }

    }
    int max = 0, ii;
    for(int i = 1; i < str.size(); i++)
    {
        if(p[i] > max)
        {
            ii = i;
            max = p[i];
        }
    }

    max--;

    string s;
    int start = ii - max ;
    int end = ii + max;
    for(int i = start; i <= end; i++)
    {
        if(str[i] != '#')
        {
            s += str[i];
        }
    }
   // cout << endl;

    delete  p;
    return s;
}