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".
Solution:Dp
思路:
dp[i][j]: the longest palindromic subsequence’s length of substring(i, j)
递推:
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])
Initialization: dp[i][i] = 1
Time Complexity: O(N) Space Complexity: O(N)
Solution Code:
public class Solution {
public int longestPalindromeSubseq(String s) {
int len = s.length();
int[][] dp = new int[len][len];
for (int i = 0; i < len; i++) {
dp[i][i] = 1;
}
for (int d = 1; d < len; d++) {
for (int i = 0; i < len - d; i++) {
int j = i + d;
if (s.charAt(i) == s.charAt(j)) {
dp[i][j] = 2 + dp[i + 1][j - 1];
}
else dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]);
}
}
return dp[0][len-1];
}
}
https://discuss.leetcode.com/topic/78603/straight-forward-java-dp-solution