5. Longest Palindromic Substring（求最長迴文子字符串）

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"

題目大意：給定一個字符串，找到最長的迴文子串。

解法一：

解題思路：迴文字符串有兩種情況：長度爲奇數，或者長度爲偶數。現在分別以這兩種情況進行解題，兩個int類型的指針left和right，分別向兩側延伸，以找到長度最大的迴文子字符串。

解題代碼：（44ms，beats 44.34%）

class Solution {
    public String longestPalindrome(String s) {
        if (s.length() <= 1) {
			return s;
		}
		int left, right;
		String res = s.substring(0, 1);
		for (int i = 0; i < s.length(); i++) {
			left = right = i;
			while (left > -1 && right < s.length() && s.charAt(left) == s.charAt(right)) {
				left--;
				right++;
			}
			if (right - left - 1 > res.length()) {
				res = s.substring(left + 1, right);
			}
		}

		for (int i = 0; i < s.length() - 1; i++) {
			left = i;
			right = i + 1;
			while (left > -1 && right < s.length() && s.charAt(left) == s.charAt(right)) {
				left--;
				right++;
			}
			if (right - left - 1 > res.length()) {
				res = s.substring(left + 1, right);
			}
		}

		return res;
    }
}

解法二：

解題思路：動態規劃

解題代碼：（180ms，beats 1.67%）

class Solution {
    public String longestPalindrome(String s) {
        int sLen = s.length();
		if (sLen <= 1) {
			return s;
		}
		boolean[][] isPalindrome = new boolean[sLen][sLen];
		// 初始化，i=j時只有一個字符，是迴文字符串。i>j時是空串，也視爲迴文字符串。i<j時先默認初始化爲false。
		for (int i = 0; i < sLen; i++) {
			for (int j = 0; j < sLen; j++) {
				if (i < j) {
					isPalindrome[i][j] = false;
				} else {
					isPalindrome[i][j] = true;
				}
			}
		}
		int left = 0, right = 0;
		for (int step = 1; step < sLen; step++) {
			for (int i = 0; i + step < sLen; i++) {
				int j = i + step;
				if (s.charAt(i) != s.charAt(j)) {
					isPalindrome[i][j] = false;
				} else {
					isPalindrome[i][j] = isPalindrome[i + 1][j - 1];
					if (isPalindrome[i][j] == true) {
						if (j - i > right - left) {
							left = i;
							right = j;
						}
					}
				}

			}
		}
		return s.substring(left, right + 1);
    }
}