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

Iteration, O(n^2) time, O(1) space

更新winIndex十分容易出錯集峦，注意evenPalinLen = 0時winStart = i + 1, winEnd = i呻澜。

class Solution {
    public String longestPalindrome(String s) {
        if (null == s) {
            return null;
        }
        
        int winStart = 0;
        int winEnd = 0;
        int len = s.length();
        
        for (int i = 0; i < len; ++i) {
            int oddPalinLen = extendPalindrome(s, i, i);
            int evenPalinLen = extendPalindrome(s, i, i + 1);
            
            if (oddPalinLen >= evenPalinLen && winEnd - winStart + 1 < oddPalinLen) {
                winStart = i - (oddPalinLen - 1) / 2;
                winEnd = i + (oddPalinLen - 1) / 2;
            } else if (oddPalinLen < evenPalinLen && winEnd - winStart + 1 < evenPalinLen) {
                winStart = i - evenPalinLen / 2 + 1;    // important!
                winEnd = i + evenPalinLen / 2 ;
            }
        }
        
        return s.substring(winStart, winEnd + 1);
    }
    
    public int extendPalindrome(String s, int i, int j) {
        while (i >= 0 && j < s.length() && s.charAt(i) == s.charAt(j)) {
            --i;
            ++j;
        }
        
        return j - i - 1;
    }
}

DP, O(n^2) time, O(n) space

dp[i][j]表示s.substring(i, j + 1)是否為palindrome渊迁。
dp[i][j] = s.charAt(i) == s.charAt(j) && (j - i < 3 || dp[i + 1][j - 1])
注意i要從大往小計算！

class Solution {
    public String longestPalindrome(String s) {
        if (s == null) return null;
        
        int len = s.length();
        boolean[][] dp = new boolean[len][len];
        int winStart = 0;
        int winEnd = 0;
        
        for (int i = len - 1; i >= 0; --i) {    // important!
            for (int j = i; j < len; ++j) {
                dp[i][j] = s.charAt(i) == s.charAt(j) && (j - i < 3 || dp[i + 1][j - 1]);
                if (dp[i][j] && j - i > winEnd - winStart) {
                    winStart = i;
                    winEnd = j;
                }
            }
        }
        
        return s.substring(winStart, winEnd + 1);
    }
}

也可以按照step去DP，保證短的substring在長的substring之前就被處理完了植榕。感覺這樣的寫法比上面這種更好理解。

class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() < 2) {
            return s;
        }
        
        int n = s.length();
        boolean[][] isPalindrome = new boolean[n][n];
        int longestStart = 0;
        int longestEnd = -1;
        
        for (int step = 0; step < n; ++step) {
            for (int i = 0; i < n - step; ++i) {
                int j = i + step;   // s.substring(i, j + 1)

                if (s.charAt(i) == s.charAt(j) && (step < 3 || isPalindrome[i + 1][j - 1])) {
                    isPalindrome[i][j] = true;

                    if (step > longestEnd - longestStart) {
                        longestStart = i;
                        longestEnd = j;
                    }
                }
            }
        }
        
        return longestEnd == -1 ? "" : s.substring(longestStart, longestEnd + 1);
    }
}