[LeetCode] Longest Palindromic Subsequence 最长回文子序列

声明：原题目转载自LeetCode，解答部分为原创

Problem ：

        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".

Solution：

        思路：动态规划问题，假定f（left, right）为字符数组string[ left， right ]中存在的回文子序列的最大值。其状态转换方程为：

        if left > right, then f( left, right ) = 0,

        else if left = right, then f( left, rigth ) = 1

         else if s[ left ] = s[ right ], then f( left, right ) = f( left + 1, right - 1 ) + 2,

        else, f( left, right ) = max( f( left + 1, right ), f( left, right - 1) )

        代码如下：

#include<iostream>#include<vector>using namespace std;class Solution {public:    int longestPalindromeSubseq(string s) {        vector<vector<int> > length(s.size(), vector<int>(s.size()));                for(int i = s.size() - 1 ; i >= 0; i --)        {        length[i][i] = 1;        for(int k = 0 ; k < i ; k ++){length[i][k] = 0;}         for(int j = i + 1; j < s.size(); j ++)        {        if(s[i] == s[j])        length[i][j] = length[i + 1][j - 1] + 2;        else        length[i][j] = max(length[i + 1][j], length[i][j - 1]);}}return length[0][s.size() - 1];    }    /*    int length(string s, int i, int j)    {    if(i == j)    return 1;    else if(i > j)        return 0;    else if(s[i] == s[j])    return length(s, i + 1, j - 1) + 2;    else    return max(length(s, i + 1, j), length(s, i, j - 1));}*/};int main(){Solution text;cout << text.longestPalindromeSubseq("aabcda") << endl;  return 0;}