[leetcode] 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, and there exists one unique longest palindromic substring.

https://oj.leetcode.com/problems/longest-palindromic-substring/

思路1（naive approach）：依次检查所有的子串（n^2），判断是否是palindrome（n），复杂度 O(n^3)。

思路2（dp）：dp[i][j] 代表从i到j的子串是否是palindrome。自下而上自左而右计算dp数组。时空复杂度都是 O(n^2)。

    dp[i][j]=1  if:

1.     i=j；
2.     s.charAt(i)==s.charAt(j)    &&    j-i<2
3.     s.charAt(i)==s.charAt(j)    &&    dp[i+1][j-1]==1
   

思路3：遍历字符串的每个字符，从这个字符出发（或者这个字符和下一个字符出发）向两侧辐射找出最长的子串。时间复杂度 O(n^2)，空间复杂度O(1)。

思路4：Manacher's algorithm。时空复杂度O(n)。

http://www.felix021.com/blog/read.php?2040

DP代码：

public class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0)
            return null;
        int start = 0;
        int end = 0;
        int len = 0;
        boolean[][] dp = new boolean[s.length()][s.length()];
        for (int i = s.length() - 1; i >= 0; i--) {
            for (int j = i; j < s.length(); j++) {
                if (i == j || (s.charAt(i) == s.charAt(j) && j - i < 2)
                        || (s.charAt(i) == s.charAt(j) && dp[i + 1][j - 1])) {
                    dp[i][j] = true;
                    if (j - i + 1 > len) {
                        len = j - i;
                        start = i;
                        end = j + 1;
                    }
                }

            }
        }

        return s.substring(start, end);
    }

    public static void main(String[] args) {
        System.out.println(new Solution().longestPalindrome("ababadccd"));
        System.out.println(new Solution().longestPalindrome("a"));
        System.out.println(new Solution().longestPalindrome(""));

    }

}

第二遍记录：注意dp的状态及转移函数。

public class Solution {
    public String longestPalindrome(String s) {
        if(s==null||s.length()==0)
            return "";
        int len=-1,start=0,end=0;
        boolean[][] dp = new boolean[s.length()][s.length()];
        
        for(int i=s.length()-1;i>=0;i--){
            for(int j=i;j<s.length();j++){
                if(i==j||(s.charAt(i)==s.charAt(j)&&(j-i<=2||dp[i+1][j-1]))){
                    dp[i][j]=true;
                    if(j-i>len){
                        start=i;
                        end=j;
                        len=j-i;
                    }
                }
            }
            
        }
        return s.substring(start,end+1);
        
    }
}

补思路3代码：

　　注意向外扩展有两种情况。

　　注意每次匹配向外匹配时长度增加2。

public class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() <= 1)
            return s;
        int n = s.length();
        int maxLen = 0;
        String res = "";

        int tmpLen = 0;
        for (int i = 0; i < n - 1; i++) {
            tmpLen = getPalin(s, i);
            if (tmpLen > maxLen) {
                maxLen = tmpLen;
                res = s.substring(i - (tmpLen - 1) / 2, i - (tmpLen - 1) / 2 + tmpLen);
            }
            tmpLen = getPalin(s, i, i + 1);
            if (tmpLen > maxLen) {
                maxLen = tmpLen;
                res = s.substring(i - (tmpLen / 2 - 1), i - (tmpLen / 2 - 1) + tmpLen);
            }

        }
        return res;

    }

    private int getPalin(String s, int idx) {
        int len = 1;
        int i = idx - 1, j = idx + 1;
        while (i >= 0 && j < s.length()) {
            if (s.charAt(i) == s.charAt(j)) {
                len += 2;
                i--;
                j++;
            } else
                break;
        }
        return len;

    }

    private int getPalin(String s, int idx, int idy) {
        int len = 0;
        int i = idx, j = idy;
        while (i >= 0 && j < s.length()) {
            if (s.charAt(i) == s.charAt(j)) {
                len += 2;
                i--;
                j++;
            } else
                break;
        }
        return len;

    }

    public static void main(String[] args) {
        System.out.println(new Solution().longestPalindrome("abccba"));
    }

}

思路4代码：

　　注意开头结尾的符号不要设成一样

public class Solution {

    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0)
            return "";
        int n = s.length();
        // preprocess the string
        StringBuilder sb = new StringBuilder();
        sb.append("$#");
        for (int i = 0; i < n; i++) {
            sb.append(s.charAt(i));
            sb.append('#');
        }
        sb.append('*');
        s = sb.toString();
        n = s.length();
        // System.out.println(s);
        int mx = 0, id = 0;
        int[] p = new int[n];
        int res = 0;
        int resIdx = -1;
        for (int i = 1; i < n - 1; i++) {
            p[i] = mx > i ? Math.min(p[2 * id - i], mx - i) : 1;
            while (s.charAt(i + p[i]) == s.charAt(i - p[i]))
                p[i]++;
            // update mx and id
            if (i + p[i] > mx) {
                mx = i + p[i];
                id = i;
            }
            // record the max of p[i];
            if (p[i] > res) {
                res = p[i];
                resIdx = i;
            }
        }
        // construct the result string
        String resStr = s.substring(resIdx - (res - 1), resIdx + (res - 1) + 1);
        StringBuilder sb2 = new StringBuilder();
        for (int i = 0; i < resStr.length(); i++)
            if (resStr.charAt(i) != '#')
                sb2.append(resStr.charAt(i));
        resStr = sb2.toString();
        return resStr;
    }

    public static void main(String[] args) {
        System.out.println(new Solution().longestPalindrome("12212321"));
    }

}

参考：

http://www.programcreek.com/2013/12/leetcode-solution-of-longest-palindromic-substring-java/

http://www.cnblogs.com/TenosDoIt/p/3675788.html

http://blog.csdn.net/worldwindjp/article/details/22066307