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  • [leetcode-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"

    思路:

    参考:

    动态规划解决。设立一个len行len列的dp数组。dp[i][j]表示字符串i~j下标所构成的子串中最长回文子串的长度。最后我们需要返回的是dp[0][len-1]的值。
    dp数组这样更新:首先i指针从尾到头遍历,j指针从i指针后面一个元素开始一直遍历到尾部。一开始dp[i][i]的值都为1,如果当前i和j所指元素相等,说明能够加到i~j的回文子串的长度中,所以更新dp[i][j] = dp[i+1][j-1] + 2; 如果当前元素不相等,那么说明这两个i、j所指元素对回文串无贡献,则dp[i][j]就是从dp[i+1][j]和dp[i][j-1]中选取较大的一个值即可。

    https://www.liuchuo.net/archives/3204

    int longestPalindromeSubseq(string s)
    {
      int n = s.size();
      vector<vector<int>>dp(n,vector<int>(n));
      
      for(int i=n-1;i>=0;i++)
      {
        dp[i][i] = 1;
        for(int j = i+1;j<n;j++)
        {
          if(s[i] == s[j]) dp[i][j] = dp[i+1][j-1]+2;
          else  dp[i][j] = max(dp[i+1][j] , dp[i][j-1]);       
        }    
      }  
      return dp[0][n-1];
    }
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  • 原文地址:https://www.cnblogs.com/hellowooorld/p/6986036.html
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