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];
}
};