[LeetCode][DP] 5. Longest Palindromic Substring.md

Description

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

Example 1:

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

Example 2:

Input: "cbbd"
Output: "bb"

Solution

Approach 4: Expand Around Center

解题思路 依次以某一个字符或两个字符为中心点，向两边扩展，寻找最大回文子串。需要特别注意回文的字符是偶数个和奇数个的情况。 所以这里的寻找回文子串的函数 expandAroundCenter  有两个参数作为中心，分别处理上述两种情况。

  class Solution {
	public String longestPalindrome(String s) {
        if (s == null || s.length() < 1) return "";
        int start = 0, end = 0;
        for (int i = 0; i < s.length(); i++) {
            int len1 = expandAroundCenter(s, i, i);
            int len2 = expandAroundCenter(s, i, i + 1);
            int len = Math.max(len1, len2);

            // 如果是以某个字符为中心的回文子串最长（len1），len一定是奇数（babad）
            // 如果是以两个字符为中心的回文子串最长（len2），len一定是偶数（cbbd）
            if (len > end - start + 1) {
                // 0    1   2   3   4
                // b    a   b   a   d
                // i = 2时，得到的len = 3，start = 1, end = 3
                // c    b   b   d
                // i = 1时，得到的len=2，start = 1, end = 2
                start = i - (len - 1) / 2;
                end = i + len / 2;
            }
        }
        return s.substring(start, end + 1);
    }

    /**
     * 以left和right为中心，向两边扩展去找最长回文子串
     *
     * @param s
     * @param left
     * @param right
     * @return
     */
    private int expandAroundCenter(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++;
        }
        // 这里因为最后一次循环L--了R++，所以最后的长度需要减1
        return R - L - 1;
    }
 }

Approach 3: Dynamic Programming

To improve over the brute force solution, we first observe how we can avoid unnecessary re-computation while validating palindromes. Consider the case "ababa". If we already knew that "bab" is a palindrome, it is obvious that "ababa" must be a palindrome since the two left and right end letters are the same.

We define  as following:

Therefore, _ and _

The base cases are:

This yields a straight forward DP solution, which we first initialize the one and two letters palindromes, and work our way up finding all three letters palindromes, and so on...

Complexity Analysis

• Time complexity : O(n^2). This gives us a runtime complexity of O(n^2).
• Space complexity : O(n^2). It uses O(n^2) space to store the table.

Additional Exercise Could you improve the above space complexity further and how?

i, j	0(b)	1(a)	2(b)	3(a)	4(d)
0(b)	true	f	t	f	f
1(a)	---	true	f	t	f
2(b)	---	---	true	f	f
3(a)	---	---	---	true	f
4(d)	---	---	---	---	true

如图所示， P(i, i) 肯定为 true 。

1. 我们只需要判断 j > i  的情况。
2. **对于 **i** **j** 的遍历，一定是从  **i: n -> 0** **j: i -> n** 的顺序，因为判断 **P(i, j)** 需要知道 **P(i +1, j-1)** 的结果。如图所示 **P(1, 4) = P(2, 3) and S_i == S_j** ，要先算出 **P(2, 3)** **
3. 需要特殊处理 P(i, i+1) 的情况，该情况下只要满足 S_i == S_i+1 就为 true 

class Solution {  
	public String longestPalindrome(String s) {
        if (s == null || s.length() < 1) return "";
        int n = s.length();
        int start = 0, end = 0;
        boolean[][] p = new boolean[n][n];
		
        // i从大到小，j从小到大遍历
        for (int i = n - 2; i >= 0; i--) {
            for (int j = i + 1; j < n; j++) {
                // 这里并没有在一开始初始化p(i,i)的值为true，
                // 而是直接在这里通过 j - i <= 2 的条件和P(i, i+1)的特殊情况一起处理了
                if (s.charAt(i) == s.charAt(j) && (j - i <= 2 || p[i + 1][j - 1])) {
                    p[i][j] = true;
                    if (j - i + 1 > end - start + 1) {
                        start = i;
                        end = j;
                    }
                }
            }
        }
        return s.substring(start, end + 1);
    }
}