[LeetCode] Shortest Palindrome

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.

For example:

Given "aacecaaa", return "aaacecaaa".

Given "abcd", return "dcbabcd".

Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Thanks to @Freezen for additional test cases.

https://leetcode.com/problems/shortest-palindrome/

其实问题能够转化成求从s[0]开始的最长回文，找到以s[0]开始的最长回文，将剩下的部分倒序补在字符串前面就是答案。在找以s[0]开始的回文的时候，能够从后向前枚举回文结束的字符，在第一个找到的地方退出便可。最坏状况下时间复杂度为O(n^2)，用C语言提交AC，704ms。

 1 bool isPalindrom(char* s, int start, int end) {
 2     while (start < end) {
 3         if (s[start] != s[end]) return false;
 4         ++start; --end;
 5     }
 6     return true;
 7 }
 8 
 9 char* shortestPalindrome(char* s) {
10     int pos = strlen(s) - 1;
11     for (; pos > 0; --pos) {
12         if (s[pos] == s[0] && isPalindrom(s, 0, pos)) break;
13     }
14     char* res = (char*) malloc(2 * strlen(s) - pos);
15     int idx = 0;
16     for (int i = strlen(s) - 1; i > pos; --i) res[idx++] = s[i];
17     strcpy(res + idx, s);
18     return res;
19 }

可是用C++提交后超时，可能这并非最佳的答案吧，想到更好的方法后再回来补上代码（已经想到了，见后文）。

 1 class Solution {
 2 public:
 3     bool isPalindrom(string &s, int start, int end) {
 4         while (start < end) {
 5             if (s[start] != s[end]) return false;
 6             ++start; --end;
 7         }
 8         return true;
 9     }
10     string shortestPalindrome(string s) {
11         int pos = s.length() - 1;
12         if (pos == 0) return s;
13         for (; pos > 0; --pos) {
14             if (s[pos] == s[0] && isPalindrom(s, 0, pos)) break;
15         }
16         string res;
17         for (int i = s.length() - 1; i > pos; --i) res.push_back(s[i]);
18         res += s;
19         return res;
20     }
21 };

 

求最长回文的Manacher算法时间复杂度为O(n)，能够用在这里，可是在判断最长回文的时候须要加一个条件，就是只比较那些以s[0]开始的回文，而根据Manacher算法的特性，id == p[id]时回文必定是从s[0]开始的。因此下面也是用C++写能够AC的代码，由于时间复杂度是O(n)，因此只用了16ms。关于Manacher算法，能够看我另外一篇文章：

http://www.cnblogs.com/easonliu/p/4454213.html

 1 class Solution {
 2 public:
 3     int longestPalindrom(string s) {
 4         string s1;
 5         s1.resize(2 * s.length() + 2);
 6         int idx = 0;
 7         s1[idx++] = '$';
 8         s1[idx++] = '#';
 9         for (char a : s) {
10             s1[idx++] = a;
11             s1[idx++] = '#';
12         }
13         vector<int> p(s1.length(), 0);
14         int res = 0;
15         for (int id = 0, i = 1; i < s1.length(); ++i) {
16             if (i < id + p[id]) p[i] = min(p[2 * id - i], id + p[id] - i);
17             else p[i] = 1;
18             while (s1[i + p[i]] == s1[i - p[i]]) ++p[i];
19             if (id + p[id] < i + p[i]) id = i;
20             if (p[i] == i) res = max(res, i);
21         }
22         return res - 1;
23     }
24     
25     string shortestPalindrome(string s) {
26         int pos = longestPalindrom(s);
27         string res;
28         for (int i = s.length() - 1; i >= pos; --i) res.push_back(s[i]);
29         return res + s;
30     }
31 };