leetcode解题: Longest Palindromic Substring (5)

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Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

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Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example:

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Input: "cbbd"
Output: "bb"

解法1：DP: O(N^2)

字符串的问题，如果牵涉到substring(i,j)的，要是往dp方向上考虑就需要用二维的。这题也不例外。
dp[i][j]表示的是，(i,j)这个字符串是否是palindrome。
有三种情况可以考虑，一个是i == j, 一个是相邻，还有一个是距离大于2。
每次找到一个palindrome的时候都更新下max，同时也更新一下res的string。
要注意的是我们要求矩阵从下往上搜索，并且j >= i, 所以我们只搜索了上半部的矩阵。
Java

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public class Solution {
    public String longestPalindrome(String s) {
        int n = s.length();
        boolean[][] state = new boolean[n][n];
        int maxlen = Integer.MIN_VALUE;
        String res = "";
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                if (i == j) {
                    state[i][j] = true;
                } else if (j == i + 1) {
                    state[i][j] = s.charAt(i) == s.charAt(j);
                } else {
                    state[i][j] = s.charAt(i) == s.charAt(j) && state[i + 1][j - 1];
                }
                if (state[i][j]) {
                    int len = j - i + 1;
                    if (len > maxlen) {
                        maxlen = len;
                        res = s.substring(i, j + 1);
                    }
                }
            }
        }
        return res;
    }
}

解法2：O(n^2) Time, O(1) Space

Java

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class Solution {
    public String longestPalindrome(String s) {
        if (s == null) {
            return s;
        }
        // loop through each possible expanding center
        int start = 0, end = 0;
        int maxLen = Integer.MIN_VALUE;
        for (int i = 0; i < s.length(); i++) {
            int len1 = expand(s, i, i);
            int len2 = expand(s, i, i + 1);
            int len = Math.max(len1, len2);
            if (len > end - start) {
                start = i - (len - 1) / 2;
                end = i + len / 2;
            }
        }
        return s.substring(start, end + 1);
    }
    private int expand(String s, int left, int right) {
        int L = left, R = right;
        while (L >= 0 && R < s.length() && s.charAt(L) == s.charAt(R)) {
            L--;
            R++;
        }
        return R - L - 1;
    }
}