动态规划——516. Longest Palindromic Subsequence[Medium]
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题目描述
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:
4One possible longest palindromic subsequence is "bbbb".
Example 2:
Input:
"cbbd"Output:
2One possible longest palindromic subsequence is "bb".
注意子串和子序列的区别,子串是连续的,子序列要按顺序但不一定连续。本题要找最长的回文子序列。
解题思路
1)数组result[i][j]表示字符串i~j之间的最长回文子序列。
2)判断str[i]== str[j],是的话,result[i][j] = result[i+1][j-1]+2,不然
result[i][j]= max(result[i+1][j],result[i][j-1])
3)返回result[0][str。size()-1]
注意:
1)因为result[i][j]可能等于result[i+1][j-1],所以,i+1必须在 i 之前算出来,j-1必须在 j 之前算出来。因此 i 逆序遍历,j 顺序
2)这是动态规划算法,result[i][j]要依赖于之前的结果,所以str[i]== str[j]时,不能简单的result[i][j]++,要
result[i][j] = result[i+1][j-1]+2
代码如下
class Solution {public: int longestPalindromeSubseq(string s) {vector<int> tmp(s.size(), 0);vector<vector<int> > result(s.size(), tmp);for (int i = 0; i < s.size(); i++)result[i][i] = 1;for (int i = s.size()-1; i >= 0; i--){for (int j = i + 1; j < s.size(); j++) {if (s[i] == s[j]) {result[i][j] = result[i + 1][j - 1] + 2;}else {result[i][j] = fmax(result[i][j - 1], result[i + 1][j]);}}}return result[0][s.size() - 1];}};
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