【LeetCode】005. Longest Palindromic Substring

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"

题解：

Solution 1

　　暴力搜索，所有可能，注意到"bab"和"baab"两种类型的回文字符串即可。

class Solution {
public:
    string longestPalindrome(string s) {
        int start = 0, len = 0;
        int n = s.size();
        if (n < 2)
            return s;
        for (int i = 0; i < n - 1; ++i) {
            rangeOfPalindrome(s, i, i + 1, start, len); // "baab"
            rangeOfPalindrome(s, i, i, start, len);   // "bab"
        }
        return s.substr(start, len);
    }
    
    void rangeOfPalindrome(const string& s, int left, int right, int& start, int& len) {
        int length = len;
        int step = 0;
        while ((left - step >= 0) && (right + step < s.size())) {
            if (s[left - step] != s[right + step])
                break;
            ++step;
        }
        length = 2 * (step - 1) + right - left + 1;
        if (length > len) {
            len = length;
            start = left - (step - 1);
        }
    }

};

 　　代码简化：

class Solution {
public:
    string longestPalindrome(string s) {
        string res;
        int n = s.size(), idx = 0, len = 0;
        for (int i = 0; i < n; ++i) {
            search(s, i, i, idx, len);
            search(s, i, i + 1, idx, len);
        }
        return s.substr(idx, len);
    }
    void search(const string& s, int i, int j, int& idx, int& len) {
        while (i >= 0 && j < s.size() && s[i] == s[j]) {
            --i;
            ++j;
        }
        if (len < j - i - 1) {
            len = j - i - 1;
            idx = i + 1;
        }
    }
};

Soluion 2

　　DP问题。一个长度为 n（n>1) 的回文字符串S（s1, s2,...,sn)，若将字符s0和sn+1分别放置在S的首尾，此时如果s0 == sn+1,那么新的字符串S'也一定是回文字符串。

　　那么递归式为 dp[i][j] = 1                                            if i == j

　　　　　　　　　　　= s[i] == s[j]　                           if i + 1 = j

                                        = s[i] == s[j] && dp[i + 1][j - 1]   if i + 1 < j

class Solution {
public:
    string longestPalindrome(string s) {
        int n = s.size();
        if (n < 1)
            return s;
        int len = 1, start = 0;
        vector<vector<int>> dp(n, vector<int>(n, 0));
        
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j <= i; ++j) {
                dp[j][i] = (s[i] == s[j]) && (i - j <= 1 || dp[j + 1][i - 1]);
                if (dp[j][i] && len < i - j + 1) {
                    len = i - j + 1;
                    start = j;
                }
            }
        }
        
        return s.substr(start, len);
    }
};

　　Manacher算法

Solution 3