[LeetCode]Longest Palindromic Subsequence

题目描述:

LeetCode 516. Longest Palindromic Subsequence

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:

求最长回文子序列的长度

解题思路:

解法I 动态规划(Dynamic Programming)

状态转移方程:

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

上式中,dp[i][j]表示s[i .. j]的最大回文子串长度

Java代码:

public class Solution {
    public int longestPalindromeSubseq(String s) {
        int size = s.length();
        int[][] dp = new int[size][size];
        for (int i = size - 1; i >= 0; i--) {
            dp[i][i] = 1;
            for (int j = i + 1; j < size; 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][size - 1];
    }
}

解法II 动态规划(Dynamic Programming)

问题转化为求s与reversed(s)的最长公共子序列

令s' = reversed(s), size = len(s)

dp[i][j]表示s[0 .. i]与s'[0 .. j]的最长公共子序列的长度

枚举回文串的中点m,求dp[m][size - m] * 2 以及 dp[m - 1][size - m] * 2 + 1的最大值

Java代码:

public class Solution {
    public int longestPalindromeSubseq(String s) {
        int size = s.length();
        int[][] dp = new int[size + 1][size + 1];
        for (int i = 1; i <= size; i++) {
            for (int j = 1; j <= size; j++) {
                if (s.charAt(i - 1) == s.charAt(size - j)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
                }
            }
        }
        int ans = s.length() > 0 ? 1 : 0;
        for (int m = 0; m < size; m++) {
            ans = Math.max(dp[m][size - m] * 2, ans);
            if (m > 0) ans = Math.max(dp[m - 1][size - m] * 2 + 1, ans);
        }
        return ans;
    }
}

 

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