分两个关系，如果s[i]==s[j]，table[i][j] = table[i+1][j-1]+2，如果s[i] != s[j]，table[i][j] = max(table[i+1][j],table[i][j-1]。
边界条件i==j，只有一个字符,table[i][j]=1，i+1==j，只有两个字符，table[i][j]=s[i]==s[j]?2:1。
class Solution { public: int longestPalindromeSubseq(string s) { const int tableSize = int(s.size()); if (tableSize == 0) { return 0; } int table[tableSize][tableSize]; for (int len = 0; len < tableSize; len++) { for (int i = 0; i < tableSize - len; i++) { if (len == 0) { table[i][i] = 1; } else if (len == 1) { table[i][i + 1] = s[i] == s[i + 1]?2:1; } else if (s[i] == s[i + len]) { table[i][i + len] = 2 + table[i + 1][i + len - 1]; } else { table[i][i + len] = max(table[i + 1][i + len], table[i][i + len - 1]); } } } return table[0][tableSize - 1]; } };
