Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example:
Input: "babad" Output: "bab" Note: "aba" is also a valid answer.
Example:
Input: "cbbd" Output: "bb"
题解：
Solution 1
暴力搜索，所有可能，注意到"bab"和"baab"两种类型的回文字符串即可。
1 class Solution { 2 public: 3 string longestPalindrome(string s) { 4 int start = 0, len = 0; 5 int n = s.size(); 6 if (n < 2) 7 return s; 8 for (int i = 0; i < n - 1; ++i) { 9 rangeOfPalindrome(s, i, i + 1, start, len); // "baab" 10 rangeOfPalindrome(s, i, i, start, len); // "bab" 11 } 12 return s.substr(start, len); 13 } 14 15 void rangeOfPalindrome(const string& s, int left, int right, int& start, int& len) { 16 int length = len; 17 int step = 0; 18 while ((left - step >= 0) && (right + step < s.size())) { 19 if (s[left - step] != s[right + step]) 20 break; 21 ++step; 22 } 23 length = 2 * (step - 1) + right - left + 1; 24 if (length > len) { 25 len = length; 26 start = left - (step - 1); 27 } 28 } 29 30 };
Soluion 2
DP问题。一个长度为 n（n>1) 的回文字符串S（s1, s2,...,sn)，若将字符s0和sn+1分别放置在S的首尾，此时如果s0 == sn+1,那么新的字符串S‘也一定是回文字符串。
那么递归式为 dp[i][j] = 1 if i == j
= s[i] == s[j] if i + 1 = j
= s[i] == s[j] && dp[i + 1][j - 1] if i + 1 < j
1 class Solution { 2 public: 3 string longestPalindrome(string s) { 4 int n = s.size(); 5 if (n < 1) 6 return s; 7 int len = 1, start = 0; 8 vector<vector<int>> dp(n, vector<int>(n, 0)); 9 10 for (int i = 0; i < n; ++i) { 11 for (int j = 0; j <= i; ++j) { 12 dp[j][i] = (s[i] == s[j]) && (i - j <= 1 || dp[j + 1][i - 1]); 13 if (dp[j][i] && len < i - j + 1) { 14 len = i - j + 1; 15 start = j; 16 } 17 } 18 } 19 20 return s.substr(start, len); 21 } 22 };
Manacher算法
Solution 3
