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;
}
}
本文链接:http://bookshadow.com/weblog/2017/02/12/leetcode-longest-palindromic-subsequence/
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