516. Longest Palindromic Subsequence

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".

本题应该使用DP的方法来做，先考虑DP是需要用数组来保存数据还是一个整型来保存数据，注意本题的特点，本题是考察回文，因此需要左端点和右端点，因此需要用到二维数组来解决问题。然后考虑状态方程，考虑到：if(charAt(i)==charAt(j)) dp[i][j] = dp[i+1][j-1]+2;如果不相等，则取比他范围小的回文最大子字符串，即：dp[i][j] = Math.max(dp[i+1][j],dp[i][j-1]);最后考虑初值情况，本题初值是dp[i][i] = 1;可以作为已知，代码如下：

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

DP还可以使用自顶向下来做，代码如下：

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