Given a string s, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.
Example 1:
Input:"aacecaaa"
Output:"aaacecaaa"
Example 2:
Input:"abcd"
Output:"dcbabcd"
最初的思路就是从头开始增加0,1,2,...len-1个字母,然后判断是否回文,但是最后一个例子会超时 class Solution { public: bool isPalindrome(string s, int l, int r){ for(int i = l, j = r; i <= j; i++, j--){ if(s[i] != s[j]) return false; } return true; } string shortestPalindrome(string s) { int len = s.length(); string res(s); int num = 0; for(int i = 0; i < len; i++){ if(isPalindrome(s, 0, len - i - 1)){ num = i; break; } } for(int i = 0; i < num; i++){ res.insert(0, 1, s[len - num + i]); } return res; } };
动态规划就是找到开始最长的回文字符串,然后把后半段补在前面,比如aaacecaa 开始最长的回文是aaa, 那么只要在前面补上aacec即可 但是最后一个案例也要超时,ccc class Solution { public: string shortestPalindrome(string s) { int len = s.length(); string res(s); int maxlen = 1; vector<vector<bool> > dp(len, vector<bool>(len, false)); for(int i = 0; i < len; i++) dp[i][i] = true; for(int i = 0; i < len - 1; i++){ if(s[i] == s[i + 1]) dp[i][i + 1] = true; if(dp[0][1]) maxlen = 2; } for(int l = 3; l <= len; l++){ for(int j = 0; j < len - l + 1; j++){ int k = j + l - 1; if(s[j] == s[k] && dp[j + 1][k - 1]) dp[j][k] = true; } if(dp[0][l - 1]) maxlen = l; } for(int p = maxlen; p < len; p++) res.insert(0, 1, s[p]); return res; } };