[leetCode刷题笔记]516. Longest Palindromic Subsequence

动态规划解决问题，dp[][]存放string（i，j）之间palim长度。

public class Solution {
    public int longestPalindromeSubseq(String s) {
        int[][] dp = new int[s.length()][s.length()];
        
        for (int i = s.length() - 1; i >= 0; i--) {
            dp[i][i] = 1;
            for (int j = i + 1; j < s.length(); j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = 2 + dp[i + 1][j - 1];
                }
                else {
                    dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
                }
            }
        }
        return dp[0][s.length()-1];
    }
}

递归其实思想也一样。用个二维数组来储存最大长度

public class Solution {
    public int longestPalindromeSubseq(String s) {
        return helper(s, 0, s.length() - 1, new Integer[s.length()][s.length()]);
    }
    
    private int helper(String s, int i, int j, Integer[][] memo) {
        if (memo[i][j] != null) {
            return memo[i][j];
        }
        if (i > j)      return 0;
        if (i == j)     return 1;
        
        if (s.charAt(i) == s.charAt(j)) {
            memo[i][j] = helper(s, i + 1, j - 1, memo) + 2;
        } else {
            memo[i][j] = Math.max(helper(s, i + 1, j, memo), helper(s, i, j - 1, memo));
        }
        return memo[i][j];
    }
}