[LeetCode] 516. Longest Palindromic Subsequence 最长回文子序列
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".
给一个字符串，求最大的回文子序列，子序列和子字符串不同，不需要是连续的字符。
解法：DP
State: dp[i][j], 表示[i,j]区间内的字符串的最长回文子序列。如果s[i]==s[j]，那么i和j就可以增加2个回文串的长度，我们知道中间dp[i + 1][j - 1]的值，那么其加上2就是dp[i][j]的值。如果s[i] != s[j]，那么我们可以去掉i或j其中的一个字符，然后比较两种情况下所剩的字符串谁dp值大，就赋给dp[i][j]。
Function: dp[i][j] = dp[i + 1][j - 1] + 2 if (s[i] == s[j]) or max(dp[i + 1][j], dp[i][j - 1]) if (s[i] != s[j])
C++: dp[i][j]
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];
}
};
C++: dp[i]
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;
}
};
类似题目：
[LeetCode] 125. Valid Palindrome 有效回文
[LeetCode] 9. Palindrome Number 验证回文数字
[LeetCode] 5. Longest Palindromic Substring 最长回文子串
