題目

https://leetcode.com/problems/longest-palindromic-subsequence/

題解

1、遞歸（超時）

遞歸 -> 傻緩存 ->DP

class Solution {public int longestPalindromeSubseq(String s) {if (s.length() == 0) return 0;return process(s, 0, s.length() - 1);}public int process(String s, int L, int R) {if (L < 0 || R == s.length() || L > R) return 0;if (L == R) {return 1;}int p1 = process(s, L + 1, R - 1);int p2 = process(s, L, R - 1);int p3 = process(s, L + 1, R);int p4 = s.charAt(L) == s.charAt(R) ? process(s, L + 1, R - 1) + 2 : 0;int res = Math.max(Math.max(p1, p2), Math.max(p3, p4));return res;} }

2、傻緩存（AC）

class Solution {public int longestPalindromeSubseq(String s) {if (s.length() == 0) return 0;int[][] dp = new int[s.length()][s.length()];return process(s, 0, s.length() - 1, dp);}public int process(String s, int L, int R, int[][] dp) {if (L < 0 || R == s.length() || L > R) return 0;if (dp[L][R] != 0) return dp[L][R];if (L == R) {dp[L][R] = 1;return 1;}int p1 = process(s, L + 1, R - 1, dp);int p2 = process(s, L, R - 1, dp);int p3 = process(s, L + 1, R, dp);int p4 = s.charAt(L) == s.charAt(R) ? process(s, L + 1, R - 1, dp) + 2 : 0;int res = Math.max(Math.max(p1, p2), Math.max(p3, p4));dp[L][R] = res;return res;} }

3、dp（AC）

class Solution {public int longestPalindromeSubseq(String s) {if (s.length() == 0) return 0;int N = s.length();int[][] dp = new int[N][N];// base casefor (int i = 0; i < N; i++) {dp[i][i] = 1;}// 根據遞歸填寫依賴for (int i = 0; i < N; i++) {for (int j = 0; j < N - i - 1; j++) {int L = j;int R = i + j + 1;int p1 = dp[L + 1][R - 1];int p2 = dp[L][R - 1];int p3 = dp[L + 1][R];int p4 = s.charAt(L) == s.charAt(R) ? dp[L + 1][R - 1] + 2 : 0;dp[L][R] = Math.max(Math.max(p1, p2), Math.max(p3, p4));}}return dp[0][N - 1];} }

總結

以上是生活随笔為你收集整理的leetcode 516. Longest Palindromic Subsequence | 516. 最长回文子序列（递归 -＞ 傻缓存 -＞DP）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。