Week Training: 516 Longest Palindromic Subsequence

原创于 2017-04-30 10:50:46 发布 · 200 阅读

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本文介绍了一种使用动态规划求解最长回文子序列问题的方法。通过二维数组记录子问题解，递归地更新状态，当序列两端字符相同时增加长度，否则选择较长子序列。初始状态为每个单独字符都是长度为1的回文子序列。

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The main difficulty is we have to find the length of longest palindromic subsequence, but not a substring. Which means the sequence may not be continuous. There are many ways to do, but using dynamic programming, the method is concise but hard to come up with. We first use a 2d array length[i][j] to store the length of each subsequence, with i, j means the length of longest PS between i and j in the string. The recurrence part add the length by 2 when the number left and right of the current sequence is the same and when not, the bigger one is chosen. The initial state is length[i][i]=1.

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int l = s.length();
        int length[1000][1000];
        for(int i=l-1;i>=0;i--){
            length[i][i]=1;
            for(int j=i+1;j<l;j++){
                if(s[i]==s[j])
                    length[i][j]=length[i+1][j-1]+2;
                else
                    length[i][j]=max(length[i+1][j],length[i][j-1]);
            }
        }
        return length[0][l-1];
    }
};