005. Longest Palindromic Substring

Posted 2020-08-14

（1）动态规划，时间复杂度O(N^2)，超时

 1 string longestPalindrome(string s)
 2 {
 3     if (s.size() < 2) return s;
 4     int max_len = 1;
 5     vector<vector<int>> vec(s.size(), vector<int>(s.size(), 0));
 6     for (int i = 0; i < s.size(); ++i) vec[i][i] = 1;
 7     for (int i = 0; i < static_cast<int>(s.size()) - 1; ++i) {
 8         if (s[i] == s[i + 1]) vec[i][i + 1] = 2;
 9     }
10     for (size_t len = 3; len <= s.size(); ++len) {
11         for (size_t startPos = 0; startPos < s.size() - 2; ++startPos) {
12             if (startPos + len - 1 >= s.size()) continue;
13             if (s[startPos] == s[startPos + len - 1] && vec[startPos + 1][startPos + len - 2]) {
14                 vec[startPos][startPos + len - 1] = vec[startPos + 1][startPos + len - 2] + 2;
15             }
16             else {
17                 vec[startPos][startPos + len - 1] = 0;
18             }
19         }
20     }
21     int left = 0, right = 0;
22     for (size_t i = 0; i < s.size(); ++i) {
23         for (size_t j = 0; j < s.size(); ++j) {
24             if (vec[i][j] > max_len) {
25                 max_len = vec[i][j];
26                 left = i; right = j;
27             }
28         }
29     }
30     return s.substr(left, right - left + 1);
31 }

 

（2）以当前点为中心点，枚举所有可能的回文串，时间复杂度O(N^2)，AC

 1 class Solution {
 2 public:
 3     string longestPalindrome(string s) {
 4         int max_len = 1;
 5         int left = 0, right = 0;
 6         for (int i = 0; i < static_cast<int>(s.size()); ++ i) {
 7             int len = 1;
 8             int current = 1;
 9             // odd
10             while (i - len >= 0 && i + len < s.size()) {
11                 if (s[i - len] == s[i + len]) {
12                     current += 2; ++ len;
13                 }
14                 else break;
15             }
16             if (current > max_len) {
17                 max_len = current;
18                 left = i - len + 1;
19                 right = i + len - 1;
20             }
21             // enen
22             if (i + 1 < s.size() && s[i] == s[i + 1]) {
23                 len = 1; current = 2;
24                 while (i - len >= 0 && i + len + 1 < s.size()) {
25                     if (s[i - len] == s[i + len + 1]) {
26                         current += 2; ++ len;
27                     }
28                     else  break;
29                 }
30                 if (current > max_len) {
31                     max_len = current;
32                     left = i - len + 1;
33                     right = i + len;
34                 }
35             }
36         }
37         if (max_len == 1) return s.substr(0, 1);
38         else return s.substr(left, right - left + 1);
39     }
40 };