516. Longest Palindromic Subsequence-优快云博客

本文链接：https://blog.youkuaiyun.com/mupengfei6688/article/details/78890622

本文介绍了解决LeetCode 516题——寻找字符串中的最长回文子序列的问题。通过动态规划的方法，给出了详细的算法实现过程及代码示例。

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516. Longest Palindromic Subsequence

题目描述：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”.
题目大意：给定一个字符串，找出字符串中最长的回文子串

思路：DP dp[i][j]表示i到j范围最长的回文串状态转移方程

dp[i][j] = dp[i+1][j-1] + 2 if s.charAt(i) == s.charAt(j)
otherwise, dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1])

代码

package DP;

/**
* @author OovEver
* 2017/12/25 11:03
*/
public class LeetCode516 {
  public int longestPalindromeSubseq(String s) {
//        dp[i][j]表示i到j范围最长的回文串
//        dp[i][j] = dp[i+1][j-1] + 2 if s.charAt(i) == s.charAt(j)
//        otherwise, dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1])
      int[][] dp = new int[s.length()][s.length()];
      for(int i=s.length()-1;i>=0;i--) {
          dp[i][i] = 1;
          for(int j=i+1;j<s.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][s.length() - 1];

  }
}