leetcode.214. Shortest Palindrome
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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"
.
首先确认一点基本知识,如果某个字符串str是回文的,那么str == reverse(str)
因此,将s逆转之后拼接在s后面,即news=s+reverse(s),该新字符串news首尾相同的部分,即为s中以s[0]为起始的最长回文子串pres
只不过这里我不用上述的遍历来做,而用类似KMP算法求next数组来做。
在KMP算法中求next数组就是s自我匹配的过程,next[i]的值就表示s[i]之前有几个元素是与s开头元素相同的。
因此,next[news.size()]的值就表示news中首尾相同的部分的长度。接下来就好做了。
注意:当next[news.size()]的值大于s.size()时,说明重复部分贯穿了s与reverse(s),应该修正为next[news.size()]+1-s.size()
class Solution {public: string shortestPalindrome(string s) { if(s == "") return s; string s2 = s; reverse(s2.begin(), s2.end()); string news = s + s2; int n = news.size(); vector<int> next(n+1); buildNext(news, next, n); if(next[n] > s.size()) next[n] = next[n] + 1 - s.size(); string pres = s.substr(next[n]); reverse(pres.begin(), pres.end()); return pres + s; } void buildNext(string& s, vector<int>& next, int n) { int k = -1; int j = 0; next[0] = -1; while(j < n) { if(k == -1 || s[j] == s[k]) { k ++; j ++; next[j] = k; } else { k = next[k]; } } } };
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