lintcode667

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

Given s = "bbbab" return 4

One possible longest palindromic subsequence is "bbbb".

 

DP。

定义：dp[i][j]为s.substring(i, j+1)中最长subsequence的长度

转移方程: if (chars[I] == chars[j]) then dp[I][j] = 2 + dp[i+1][j-1] else dp[I][j] = max(dp[i+1][j], dp[i][j-1]).

返回值: dp[0][s.length()-1]

原理：和substring的问题区别就在于可以跳着取，所以i,j可以不用都参与进来。对任何dp[i][j]你就分成ij都贡献，只i贡献，只j贡献，ij都不贡献四种情况，从而如以上转移方程，其中因为递增性省去了ij都不贡献的情况。

 

细节：

1.dp题目很多可以最后就返回dp记录的其中一项，省去自己打擂台。

2.从右下开始往上再往右走。 

 

 

1. 我的实现

public class Solution {
    /**
     * @param s: the maximum length of s is 1000
     * @return: the longest palindromic subsequence's length
     */
    public int longestPalindromeSubseq(String s) {
        // write your code here
        
        int result = 0;
        if (s == null) {
            return result;
        }
        
        int[][] longestLength = new int[s.length()][s.length()];
        
        //initialization for diagonal
        for (int i = 0; i < s.length(); i++) {
            longestLength[i][i] = 1;
            result = Math.max(result, longestLength[i][i]);
        }
        
        //initialization for the line in the right of the diagonal
        for (int i = 0; i + 1 < s.length(); i++) {
            longestLength[i][i + 1] = s.charAt(i) == s.charAt(i + 1) ? 2 : 1;
            result = Math.max(result, longestLength[i][i + 1]);
        }
        
        //for general cases
        for (int i = s.length() - 1; i >= 0; i--) {
            for (int j = i + 2; j < s.length(); j++){
                if (s.charAt(i) == s.charAt(j)) {
                    longestLength[i][j] = longestLength[i + 1][j - 1] + 2;
                } else {
                    longestLength[i][j] = Math.max(longestLength[i + 1][j], 
                                                   longestLength[i][j - 1]);
                }
                result = Math.max(result, longestLength[i][j]);
            }
        }
        
        return result;
    }
}

2.九章实现

public class Solution {
    /**
     * @param s the maximum length of s is 1000
     * @return the longest palindromic subsequence's length
     */
    public int longestPalindromeSubseq(String s) {
        // Write your code here
        int length = s.length();
        if (length == 0)
            return 0;
        int[][] dp = new int[length][length];
        for (int i = length - 1; i >= 0; i--) {
            dp[i][i] = 1;
            for (int j = i + 1; j < length; j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                } else {
                    dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[0][length - 1];
    }
}