leetcode516. Longest Palindromic Subsequence

题目

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”.

思路

这个题目跟Longest Palindromic Substring,有些类似，主要区别就是，子串必须是连续的，而子序列可以不连续。很明显依然用dp来做，只是转换方程有些差异，初始化的两个方程是类似的P（i，i）= 1，P（i，i+1）=2 （S[i]= S[i+1]）else P（i，i+1）=1 ,状态转移方程为P（i，j）=max(P(i,j-1),P(i+1,j),P(i+1,j-1)+2) (S[i]= S[j]),P（i，j）=max(P(i,j-1),P(i+1,j),P(i+1,j-1)) (S[i]!= S[j]),每个新的状态都可以由三个旧状态转换而来。

代码

class Solution {public:    int longestPalindromeSubseq(string s) {        //空串        if (s.empty())         return 0;        int n=s.length();        //table[i][j]表示s中i-j的回文子序列的最大长度。        int table[1000][1000]={0};        for (int len=1; len<=n; len++) {            for (int lhs=0; lhs+len<=s.size(); lhs++) {                int rhs = lhs+len-1;                if (lhs == rhs) {                    table[lhs][rhs] = 1;                } else if (lhs+1 == rhs) {       //以上两种情况是初始化状态                    table[lhs][rhs] = (s[lhs] == s[rhs]) ? 2 : 1;                  } else {                    int add = s[lhs] == s[rhs] ? 2 : 0;                    table[lhs][rhs] = max(max(table[lhs][rhs-1], table[lhs+1][rhs]), table[lhs+1][rhs-1] + add);                }            }        }        return table[0][n-1];    }};