题意:
给出一个字符串,求出其中最大的回文串字数
要点:
这题比较好的办法是用dp做:
- dp[i][j]表示从s[i]到s[j]是否为回文串,用0、1区分
- s[i]=s[j]时,dp[i][j]=dp[i+1][j-1];s[i]!=s[j]时,dp[i][j]=0
- 边界:dp[i][i]=1,dp[i][i+1]=(s[i]==s[i+1])?1:0
- 因为i、j如果从小到大的顺序来枚举的话,无法保证更新dp[i][j]的时候dp[i+1][j-1]已经被计算过。因此不妨考虑按照字串的长度和子串的初试位置进行枚举,即第一遍将长度为3的子串的dp的值全部求出,第二遍通过第一遍结果计算出长度为4的子串的dp的值…这样就可以避免状态无法转移的问题
#include<iostream> #include<string> #include<vector> #include<map> #include<set> #include<sstream> #include<functional> #include<algorithm> using namespace std; const int INF = 0xfffff; const int maxn = 1010; int dp[maxn][maxn]; int main() { string s; getline(cin, s); int ans = 1; int len = s.length(); for (int i = 0; i < len; i++) { dp[i][i] = 1; if (i + 1 < len&&s[i] == s[i + 1]) { dp[i][i + 1] = 1; ans = 2; } } for (int l = 3; l <= len; l++) { for (int i = 0; i+l-1 < len; i++) { int j = i + l - 1; if (s[i] == s[j] && dp[i + 1][j - 1] == 1) { dp[i][j] = 1; ans = l; } } } cout << ans << endl; return 0; }
/*以字符串中每个字符为对称轴*/ #include<iostream> #include<string> #include<vector> #include<map> #include<set> #include<sstream> #include<functional> #include<algorithm> using namespace std; const int INF = 0xfffff; const int maxn = 1005; int main() { string s; getline(cin, s); int len = 1; for (int i = 0; i < s.size(); i++) { for (int j = 0; j <= min(i, (int)(s.size() - i - 1)); j++) { string s2 = s.substr(i - j, 2 * j + 1);//字数为奇数 string s3 = s2; reverse(s3.begin(), s3.end()); if (s2 == s3&&s2.size() > len) len = s2.size(); string s4 = s.substr(i - j, 2 * j + 2);//字数为偶数 string s5 = s4; reverse(s5.begin(), s5.end()); if (s4 == s5&&s4.size() > len) len = s4.size(); } } cout << len << endl; return 0; }