状态:没有思路
重点是构造了一个dp[i][j],来表示s字符串i到j是否是一个回文串,代码如下:
class Solution {
public:
int countSubstrings(string s) {
int len = s.size();
int res = 0;
vector<vector<bool>> dp(len, vector<bool>(len, false));
for(int i = 0; i < len; ++i) dp[i][i] = true;
for(int i = len-1; i >= 0; --i){
for(int j = i; j < len; ++j){
if(s[i] == s[j]){
if(j - i <= 1){
res++;
dp[i][j] = true;
}else if(dp[i+1][j-1]){
res++;
dp[i][j] = true;
}
}
}
}
return res;
}
};
状态:查看思路后AC。
注意遍历顺序和初始化,代码如下:
class Solution {
public:
int longestPalindromeSubseq(string s) {
int len = s.size();
vector<vector<int>> dp(len, vector<int>(len, 0));
for(int i = 0; i < len; ++i) dp[i][i] = 1;
for(int i = len-1; i >= 0; --i){
for(int j = i+1; j < len; ++j){
if(s[i] == s[j]) dp[i][j] = dp[i+1][j-1] + 2;
else dp[i][j] = max(dp[i+1][j], dp[i][j-1]);
}
}
return dp[0][len-1];
}
};