[LeetCode] Longest Palindromic Substring

题目

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.

思路1

最直接的思路当然是找到每一个子串，检测是否是回文字符串。

代码1

public class Solution {    public String longestPalindrome(String s) {        int len = s.length();                if (s == null)            return "Wrong Input!";        if (len == 0)            return "";                for (int i = len;i > 0;i--) {            for (int j = 0;j <= len-i;j++) {                if (isPalindrome(s, j, j+i-1)){                    return s.substring(j, j+i);                }            }        }        return null;    }        public static boolean isPalindrome(String s, int low, int high) {        if (low >= high)            return true;                if (s.charAt(low) == s.charAt(high))            return isPalindrome(s, ++low, --high);            else            return false;    }}

显然这种做法会导致Time Limit Exceeded。 

思路2

这类问题一个常规的思路是动态规划，可以将时间复杂度降低到O(N²)，首先给出初始状态和状态转移方程。

初始状态是

P[ i, i ] ← true
P[ i, i+1 ] ← Si = Si+1

状态转移方程

P[ i, j ] ← P[ i+1, j-1 ] 且Si = Sj  

代码2

public class Solution {    public String longestPalindrome(String s) {        int len = s.length();        int beginIndex = 0, maxLen = 1;                boolean[][] state = new boolean[1000][1000];                for (int i = 0; i < len; i++) {            state[i][i] = true;        }                for (int i = 0; i < len-1; i++) {            if (s.charAt(i) == s.charAt(i+1)) {                state[i][i+1] = true;                beginIndex = i;                maxLen = 2;            }         }                for (int k = 3; k <= len; k++) {            for (int i = 0; i <= len-k ;i++) {                int j = i + k - 1;                                if ((s.charAt(i) == s.charAt(j)) && (state[i+1][j-1])) {                    state[i][j] = true;                    beginIndex = i;                    maxLen = k;                }            }        }        return s.substring(beginIndex, beginIndex+maxLen);    }}