第十四周

516. Longest Palindromic Subsequence

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• Total Accepted: 9935
• Total Submissions: 23475
• Difficulty: Medium
• Contributors:Stomach_ache

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".

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思路：

设字符串中从i到j中回文子序列的最长长度为length（i，j），由回文子序列的性质可得，如果s[i] = s[j]且i+1 <= j-1，length（i，j）=length（i+1，j-1）+ 2；而s[i] = s[j]且i+1 > j-1，length（i，j）= 2；若s[i] ！= s[j]则length（i，j）= max（length（i+1，j），length（i，j-1））。所以该字符串中回文子序列的最长长度为length（0，s.size() - 1）。

代码：

class Solution {
public:
    
    int longestPalindromeSubseq(string s) {
        int size = s.size();
        int num = 1;
        vector
     
     
      
      
       
       > length(size, vector
       
       
        
        (size, 1));
        for(int i = 1; i < size; i ++){
            for(int j = i - 1; j >= 0; j --){
                if(s[i] == s[j]){
                    if(j + 1 <= i - 1) length[j][i] = length[j + 1][i - 1] + 2;
                    else length[j][i] = 2;
                }
                else{
                    length[j][i] = max(length[j + 1][i], length[j][i - 1]);
                }
            }
        }
        
        return length[0][size - 1];
    }
};