LeetCode Solutions
class Solution {
public:
int longestPalindromeSubseq(string s) {
const int n = s.length();
// dp[i][j] := LPS's length in s[i..j]
dp.resize(n, vector<int>(n));
return lps(s, 0, n - 1);
}
private:
vector<vector<int>> dp;
int lps(const string& s, int i, int j) {
if (i > j)
return 0;
if (i == j)
return 1;
if (dp[i][j])
return dp[i][j];
if (s[i] == s[j])
dp[i][j] = 2 + lps(s, i + 1, j - 1);
else
dp[i][j] = max(lps(s, i + 1, j), lps(s, i, j - 1));
return dp[i][j];
}
};
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class Solution {
public int longestPalindromeSubseq(String s) {
final int n = s.length();
// dp[i][j] := LPS's length in s[i..j]
dp = new int[n][n];
return lps(s, 0, n - 1);
}
private int[][] dp;
private int lps(final String s, int i, int j) {
if (i > j)
return 0;
if (i == j)
return 1;
if (dp[i][j] > 0)
return dp[i][j];
if (s.charAt(i) == s.charAt(j))
dp[i][j] = 2 + lps(s, i + 1, j - 1);
else
dp[i][j] = Math.max(lps(s, i + 1, j), lps(s, i, j - 1));
return dp[i][j];
}
}
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class Solution:
def longestPalindromeSubseq(self, s: str) -> int:
# Dp(i, j) := LPS's length in s[i..j]
@functools.lru_cache(None)
def dp(i: int, j: int) -> int:
if i > j:
return 0
if i == j:
return 1
if s[i] == s[j]:
return 2 + dp(i + 1, j - 1)
return max(dp(i + 1, j), dp(i, j - 1))
return dp(0, len(s) - 1)
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