LeetCode——最長連續迴文串

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

給出一個字符串S，找到一個最長的連續迴文串。你能夠假設s的最大長度是1000。

 

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

 

Example:

Input: "cbbd"

Output: "bb"

 

動態規劃：

 1 class Solution {
 2 public:
 3     string longestPalindrome(string s) {
 4         if (s.empty()) return "";
 5         if (s.size() == 1) return s;
 6         
 7         int min_start = 0;
 8         int max_len = 1;
 9         int n = s.size();
10         for (int i = 0; i < n;) {
11             int j = i, k = i;
12             while (k < n-1 && s[k] == s[k+1]) {//k——k+1表示相等的同一個元素
13                 k++;                        
14             }
15             i = k + 1;
16             while (k < n-1 && s[k+1] == s[j-1] && j > 0) {//j-1——k+1爲對稱相等的元素
17                 ++k;
18                 --j;
19             }
20             
21             int new_len = k - j + 1;//當前迴文串大小
22             if (new_len > max_len) {
23                 min_start = j;
24                 max_len = new_len;
25             }
26         }
27         return s.substr (min_start, max_len);
28     }
29 };

View Code

 

 

?Manacher’s Algorithm算法

 用 mx 記錄以前計算的最長的迴文子串長度所能到達的最後邊界，用 id 記錄其對應的中心，能夠利用迴文子串中的迴文子串信息。

 1 class Solution {
 2 public:
 3     string longestPalindrome(string s) {
 4         string t ="$#";
 5         for (int i = 0; i < s.size(); ++i) {
 6             t += s[i];
 7             t += '#';
 8         }
 9         int p[t.size()] = {0}, id = 0, mx = 0, resId = 0, resMx = 0;
10         for (int i = 0; i < t.size(); ++i) {
11             p[i] = mx > i ? min(p[2 * id - i], mx - i) : 1;
12             while (t[i + p[i]] == t[i - p[i]]) ++p[i];
13             if (mx < i + p[i]) {
14                 mx = i + p[i];
15                 id = i;
16             }
17             if (resMx < p[i]) {
18                 resMx = p[i];
19                 resId = i;
20             }
21         }
22         return s.substr((resId - resMx) / 2, resMx - 1);
23     }
24 };

View Code