Given a string s, find the longest palindromic subsequence's length in s. You may assume that the maximum length of s is 1000.html
Example 1:
Input:java
"bbbab"
Output:算法
4
One possible longest palindromic subsequence is "bbbb".數組
Example 2:
Input:緩存
"cbbd"
Output:post
2
One possible longest palindromic subsequence is "bb".優化
這道題給了咱們一個字符串,讓咱們求最大的迴文子序列,子序列和子字符串不一樣,不須要連續。而關於迴文串的題以前也作了很多,處理方法上就是老老實實的兩兩比較吧。像這種有關極值的問題,最應該優先考慮的就是貪婪算法和動態規劃,這道題顯然使用DP更加合適。咱們創建一個二維的DP數組,其中dp[i][j]表示[i,j]區間內的字符串的最長迴文子序列,那麼對於遞推公式咱們分析一下,若是s[i]==s[j],那麼i和j就能夠增長2個迴文串的長度,咱們知道中間dp[i + 1][j - 1]的值,那麼其加上2就是dp[i][j]的值。若是s[i] != s[j],那麼咱們能夠去掉i或j其中的一個字符,而後比較兩種狀況下所剩的字符串誰dp值大,就賦給dp[i][j],那麼遞推公式以下:url
/ dp[i + 1][j - 1] + 2 if (s[i] == s[j])spa
dp[i][j] =code
\ max(dp[i + 1][j], dp[i][j - 1]) if (s[i] != s[j])
解法一:
class Solution { public: 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]; } };
咱們能夠對空間進行優化,只用一個一維的dp數組,參見代碼以下:
解法二:
class Solution { public: int longestPalindromeSubseq(string s) { int n = s.size(), res = 0; vector<int> dp(n, 1); for (int i = n - 1; i >= 0; --i) { int len = 0; for (int j = i + 1; j < n; ++j) { int t = dp[j]; if (s[i] == s[j]) { dp[j] = len + 2; } len = max(len, t); } } for (int num : dp) res = max(res, num); return res; } };
下面是遞歸形式的解法,memo數組這裏起到了一個緩存已經計算過了的結果,這樣能提升運算效率,使其不會TLE,參見代碼以下:
解法三:
class Solution { public: int longestPalindromeSubseq(string s) { int n = s.size(); vector<vector<int>> memo(n, vector<int>(n, -1)); return helper(s, 0, n - 1, memo); } int helper(string& s, int i, int j, vector<vector<int>>& memo) { if (memo[i][j] != -1) return memo[i][j]; if (i > j) return 0; if (i == j) return 1; if (s[i] == s[j]) { memo[i][j] = helper(s, i + 1, j - 1, memo) + 2; } else { memo[i][j] = max(helper(s, i + 1, j, memo), helper(s, i, j - 1, memo)); } return memo[i][j]; } };
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參考資料:
https://discuss.leetcode.com/topic/78603/straight-forward-java-dp-solution
https://discuss.leetcode.com/topic/78799/c-beats-100-dp-solution-o-n-2-time-o-n-space