[leetcode]516. Longest Palindromic Subsequence
Analysis
完了完了 真的要找不到工作了!!!!!苍了天了!!!—— [每天刷题并不难0.0]
Given a string s, find the longest palindromic subsequence’s length in s. You may assume that the maximum length of s is 1000.
动归解决~
dp[i][j]表示s中从i到j的子串中最长回文串的长度,于是状态转移方程为:
如果s[i]=s[j],则dp[i][j] += dp[i+1][j-1]+2;
否则dp[i][j] = max(dp[i+1][j], dp[i][j-1])。
Implement
class Solution {
public:
int longestPalindromeSubseq(string s) {
int len = s.size();
vector<vector<int>> dp(len+1, vector<int>(len+1));
for(int i=0; i<len; i++)
dp[i][i] = 1;
for(int i=len-1; i>=0; i--){
for(int j=i+1; j<len; 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][len-1];
}
};