leetcode Longest Palindromic Substring

Description:

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"

Solution:

1.暴力求解

时间复杂度为O（n³）。

2.根据定义

回文串的方法就是两个两个的对称验证是否相等，那么对于找回文字串的问题，就要以每一个字符为中心，像两边扩散来寻找回文串，不过要注意奇偶情况，于回文串的长度可奇可偶两种形式的回文都要搜索。

时间复杂度为O（n²）。

3.动态规划

P(i,j)={true,false,if the substring Si…Sj is a palindromeotherwise. 
P(i,j)={true,if the substring Si…Sj is a palindromefalse,otherwise. 

Therefore,

P(i, j) = ( P(i+1, j-1) \text{ and } S_i == S_j )P(i,j)=(P(i+1,j−1) and S​i​​==S​j​​)

The base cases are:

$$P(i,i)=true$$

P(i, i+1) = ( S_i == S_{i+1} )P(i,i+1)=(S​i​​==S​i+1​​)

时间复杂度O（n²）。

3.Manacher算法

马拉车算法，十分巧妙的把时间复杂度变为线性。即时间复杂度为O(n)。

参考：http://blog.youkuaiyun.com/dyx404514/article/details/42061017

代码：

class Solution {//19ms
public:
    string longestPalindrome(string s) {
        int length = s.size();
        int start = 0, end = 0;
        if (length == 0) return s;
        for (int i = 0; i < length; i++) {
            searchPalindrome(s, i, i, start, end);
            searchPalindrome(s, i, i+1, start, end);
        }
        return s.substr(start,end-start+1);
    }
    
    void searchPalindrome(string s, int i, int j, int &start, int &end) {
        int length = s.size();
        while (i >= 0 && j < length && s[i] == s[j])
            i--, j++;
        i++,j--;
        if (j - i > end - start) {
            start = i;
            end = j;
        }
    }
};

class Solution {//102ms
public:
    string longestPalindrome(string s) {
        int dp[s.size()][s.size()] = {0}, left = 0, right = 0, len = 0;
        for (int i = 0; i < s.size(); ++i) {
            for (int j = 0; j < i; ++j) {
                dp[j][i] = (s[i] == s[j] && (i - j < 2 || dp[j + 1][i - 1]));
                if (dp[j][i] && len < i - j + 1) {
                    len = i - j + 1;
                    left = j;
                    right = i;
                }
            }
            dp[i][i] = 1;
        }
        return s.substr(left, right - left + 1);
    }
};

class Solution {//3ms
public:
    string longestPalindrome(string s) {
        string t ="$#";
        for (int i = 0; i < s.size(); ++i) {
            t += s[i];
            t += '#';
        }
        int p[t.size()] = {0}, id = 0, mx = 0, resId = 0, resMx = 0;
        for (int i = 0; i < t.size(); ++i) {
            p[i] = mx > i ? min(p[2 * id - i], mx - i) : 1;
            while (t[i + p[i]] == t[i - p[i]]) ++p[i];
            if (mx < i + p[i]) {
                mx = i + p[i];
                id = i;
            }
            if (resMx < p[i]) {
                resMx = p[i];
                resId = i;
            }
        }
        return s.substr((resId - resMx) / 2, resMx - 1);
    }
};