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O(n2):
1 class Solution { 2 public: 3 string getP(string s, int start, int end) { 4 while (start >= 0 && end < s.size() && s[start] == s[end]) { 5 start--; 6 end++; 7 } 8 return s.substr(start+1, end - start - 1); 9 } 10 string longestPalindrome(string s) { 11 int len = s.size(); 12 if (len < 2) return s; 13 string result = s.substr(0, 1); 14 for (int i = 0; i < len-1; i++) { 15 string s1 = getP(s, i, i); 16 if (s1.size() > result.size()) result = s1; 17 string s2 = getP(s, i, i+1); 18 if (s2.size() > result.size()) result = s2; 19 } 20 return result; 21 } 22 };
TBA (O(n))
LeetCode – Refresh – Longest Palindromic Substring
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原文地址:http://www.cnblogs.com/shuashuashua/p/4352687.html