[LeetCode] 516. Longest Palindromic Subsequence 最长回文子序列

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

Example 1:
Input:

"bbbab"

Output:

4

One possible longest palindromic subsequence is "bbbb". 

Example 2:
Input:

"cbbd"

Output:

2

One possible longest palindromic subsequence is "bb".

给一个字符串，求最大的回文子序列，子序列和子字符串不同，不需要是连续的字符。

解法：DP

State: dp[i][j], 表示[i,j]区间内的字符串的最长回文子序列。如果s[i]==s[j]，那么i和j就可以增加2个回文串的长度，我们知道中间dp[i + 1][j - 1]的值，那么其加上2就是dp[i][j]的值。如果s[i] != s[j]，那么我们可以去掉i或j其中的一个字符，然后比较两种情况下所剩的字符串谁dp值大，就赋给dp[i][j]。

Function: dp[i][j] = dp[i + 1][j - 1] + 2 if (s[i] == s[j]) or max(dp[i + 1][j], dp[i][j - 1]) if (s[i] != s[j])

C++: dp[i][j]

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int n = s.size();
        vector<vector<int>> dp(n, vector<int>(n));
        for (int i = n - 1; i >= 0; --i) {
            dp[i][i] = 1;
            for (int j = i + 1; j < n; ++j) {
                if (s[i] == s[j]) {
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                } else {
                    dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[0][n - 1];
    }
};

C++: dp[i]

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int n = s.size(), res = 0;
        vector<int> dp(n, 1);
        for (int i = n - 1; i >= 0; --i) {
            int len = 0;
            for (int j = i + 1; j < n; ++j) {
                int t = dp[j];
                if (s[i] == s[j]) {
                    dp[j] = len + 2;
                } 
                len = max(len, t);
            }
        }
        for (int num : dp) res = max(res, num);
        return res;
    }
};

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[LeetCode] 125. Valid Palindrome 有效回文

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[LeetCode] 5. Longest Palindromic Substring 最长回文子串

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