PAT1040:Longest Symmetric String
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1040. Longest Symmetric String (25)
Given a string, you are supposed to output the length of the longest symmetric sub-string. For example, given "Is PAT&TAP symmetric?", the longest symmetric sub-string is "s PAT&TAP s", hence you must output 11.
Input Specification:
Each input file contains one test case which gives a non-empty string of length no more than 1000.
Output Specification:
For each test case, simply print the maximum length in a line.
Sample Input:Is PAT&TAP symmetric?Sample Output:
11
思路
求一个字符串的最长回文子串。
DP的思想,复杂度O(n^2)。
1.如果一个从i到j的字符串str(i,j)的子串str(i+1,j-1)为回文串,那么在str[i] == str[j]的情况下,字符串str(i,j)也是回文串。
2.每一个字符本身就是一个回文串。所以在每一个字符的基础上,根据1的条件来确定更长的回文串。
3.用一个bool数组isSym[1001][1001]来列举所有的情况,isSym[i][j]表示起始位置为i、终止位置为j的字符串是否是回文串。
4.检查是否有N个长度的回文串,更新最大长度maxlength。(1 <=N <= str.length())。
代码
#include<iostream> #include<vector> using namespace std; vector<vector<bool>> isSym(1001,vector<bool>(1001,false)); int main() { string s; getline(cin,s); int maxlength = 1; const int Length = s.size(); for(int i = 0;i < Length;i++) { isSym[i][i] = true; if(i < Length - 1 && s[i] == s[i + 1]) { isSym[i][i+1] = true; maxlength = 2; } } for(int len = 3;len <= Length;len++) { for(int i = 0;i <= Length - len;i++) { int j = i + len - 1; if(isSym[i+1][j-1] && s[i] == s[j]) { isSym[i][j] = true; maxlength = len; } } } cout << maxlength << endl; }
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