Longest Palindromic Substring[leetcode]

标签：leetcode

实现了两种方法，一种是DP，用循环做的，递归的话更简单。

string longestPalindrome(string s) {
        int n = s.size();
        bool dp[1001][1001];
        int maxl = 1;
        int maxs = 0;
        for (int i = n - 1; i >= 0; i--)
        {
            dp[i][i] = true;
            for (int j = i + 1; j < n; j++)
            {
                if (s[i] == s[j] && (j == i + 1 || dp[i + 1][j - 1]))
                {
                    dp[i][j] = true;
                    if (maxl < (j - i + 1))
                    {
                        maxl = (j - i + 1);
                        maxs = i;
                    }
                }
                else
                    dp[i][j] = false;
            }
        }
        return s.substr(maxs, maxl);
    }

另外一种是比较有名的O(n)的方法

string longestPalindrome(string s) {
        int n = s.size();
        int newn = n * 2 + 1;
        int p[2000];
        char * news = new char[newn];
        for (int i = 0; i < n; i++)
        {
            news[2 * i] = '*';
            news[2 * i + 1] = s[i];
        }
        news[2 * n] = '*';
        //s:abac
        //news:*a*b*a*c* 
        //P:   121412121
        int mx = 0;
        int mi = 0;
        int maxLen = 0;
        int maxIndex = 0;
        for (int i = 0; i < newn; i++)
        {
            if (mx > i)
                p[i] = min(p[2 * mi - i], mx - i);
            else
                p[i] = 1;
            for (;i - p[i] >= 0 && i + p[i] < newn && news[i + p[i]] == news[i - p[i]];p[i]++);
            if (p[i] + i > mx)
            {
                mx = p[i] + i;
                mi = i;
            }
            if (maxLen < p[i] - 1)
            {
                maxLen = p[i] - 1;
                maxIndex = i;
            }
        }
        string returnVal = s.substr((maxIndex - maxLen) / 2, maxLen);
        delete [] news;
        
        return returnVal;
    }

Longest Palindromic Substring[leetcode]

标签：leetcode

原文地址：http://blog.csdn.net/peerlessbloom/article/details/39268893