[开发工具][LeetCode] Longest Palindromic Substring

Description

Longest Palindromic Substring: Given a string s, return the longest palindromic substring in s.

Example:

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

Solution

The basic idea for this question is to use dynamic programming. This is because whether a string is a palindrome can be transfered from whether its substring is a palindrome. For example, if a string abcba is palindrome, its substring bcb also is a palindrome. But if a substring bcb is a palindrome, a string abcbd may not be a palindrome.

Moreover, we notice that the odd length string can be transfered from its previous odd length substring, and the even length string can be transfered from its previous even length substring. For example, abcba which length is 5 can be transfered from bcb which length is 3. Also, abccba which length is 6 can be transfered from bccb which length is 4.

Accroding to these two analyses, we can get the transfer equation: dp[i][j] = dp[i + 1][j - 1] + 2 if dp[i + 1][j - 1] is a palindrome and s[i] == s[j]. (i and j are indexes of the string s)

Finally, the last thing we should do is to record the start and end of longest palindromic substring for return.

Code

class Solution {
public:
    string longestPalindrome(string s) {
        int len = s.length();
        int start = 0, end = 0;
        
        vector<vector<int>> dp(2, vector(len, 0));
        dp[0][0] = 1;
        for(int i = 1; i < len; i++) {
            dp[1 & 1][i] = 1;
            if(s[i] == s[i - 1]) {
                dp[2 & 1][i] = 2;
                start = i - 1;
                end = i;
            }
        }
        
        for(int curLen = 3; curLen <= len; curLen++) {
            for(int i = len - 1; i >= curLen - 1; i--) {
                if((dp[curLen & 1][i - 1] == curLen - 2) && s[i - curLen + 1] == s[i]) {
                    dp[curLen & 1][i] = curLen;
                    start = i - curLen + 1;
                    end = i;
                }
            }
        }
        return s.substr(start, end - start + 1);
    }
};

Complexity

Time complexity: O(n ^ 2)

Space complexity: O(n) for dynamic programming