题目链接
题目要求:
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"
.
这道题用暴力破解是无法通过LeetCode上的测试的。
1. 法一(暴力破解)
超过时间!!!!
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
1 int findCenter(string s) 2 { 3 int center = 0; 4 int szS = s.size(); 5 for (int i = 1; i < szS; i++) 6 { 7 int k = 1; 8 while (k <= i && k < szS - i) 9 { 10 if (s[i - k] != s[i + k]) 11 break; 12 k++; 13 } 14 if (i - k == -1 && i > center) 15 center = i; 16 } 17 return center; 18 } 19 20 string char2String(char c) 21 { 22 stringstream ss; 23 string s; 24 ss << c; 25 ss >> s; 26 return s; 27 } 28 29 string shortestPalindrome(string s) { 30 // clear spaces 31 if (s.front() == ' ') 32 s.erase(0, 1); 33 if (s.back() == ' ') 34 s.pop_back(); 35 36 string ret; 37 int szS = s.size(); 38 if (szS == 0) 39 ret = ""; 40 else if (szS == 1) 41 { 42 s += s; 43 ret = s; 44 } 45 else 46 { 47 ret = s; 48 int center = findCenter(s); 49 int left = center; 50 int right = center + left + 1; 51 if (szS - center - 1 > center && s[center] == s[center + 1]) 52 { 53 right = szS; 54 } 55 for (int i = right; i < szS; i++) 56 { 57 ret.insert(0, char2String(s[i])); 58 } 59 } 60 61 return ret; 62 }
2. 法二(暴力破解)
法二思路来源:How to get the shortest palindrome of a string。重要语句摘录如下:
Just append the reverse of initial substrings of the string, from shortest to longest, to the string until you have a palindrome. e.g., for "acbab", try appending "a" which yields "acbaba", which is not a palindrome, then try appending "ac" reversed, yielding "acbabca" which is a palindrome.
法二相对法一来说更加简单容易理解,但同样超过时间!!!!
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
1 bool isPalindrome(string s) 2 { 3 int szS = s.size(); 4 for (int i = 0; i < szS / 2; i++) 5 { 6 if (s[i] != s[szS - i - 1]) 7 return false; 8 } 9 return true; 10 } 11 12 string shortestPalindrome(string s) 13 { 14 // clear spaces 15 if (s.front() == ' ') 16 s.erase(0, 1); 17 if (s.back() == ' ') 18 s.pop_back(); 19 // 20 string ret; 21 int szS = s.size(); 22 if (szS == 0) 23 { 24 ret = ""; 25 } 26 else if (szS == 1) 27 { 28 s += s; 29 ret = s; 30 } 31 else if (isPalindrome(s)) 32 { 33 ret = s; 34 } 35 else 36 { 37 for (int i = 0; i < szS; i++) 38 { 39 string tmpStr = s; 40 for (int j = szS - i - 1; j < szS; j++) 41 tmpStr.insert(tmpStr.begin(), s[j]); 42 43 if (isPalindrome(tmpStr)) 44 { 45 ret = tmpStr; 46 break; 47 } 48 } 49 } 50 return ret; 51 }
看来只能另寻他法了。。。。
3. 法三
法三参考自C++ 8 ms KMP-based O(n) time & O(n) memory solution,利用了KMP的思想。具体程序如下:
1 class Solution { 2 public: 3 string shortestPalindrome(string s) 4 { 5 string rev_s(s); 6 reverse(rev_s.begin(), rev_s.end()); 7 string combine = s + "#" + rev_s; // 防止在匹配的时候,不从‘#’后开始 8 9 int sz = combine.size(); 10 vector<int> match(sz, 0); 11 int k = 0; 12 for (int i = 1; i < sz; i++) { 13 while (k > 0 && combine[i] != combine[k]) 14 k = match[k - 1]; 15 16 if (combine[i] == combine[k]) 17 k = k + 1; 18 19 match[i] = k; 20 } 21 22 return rev_s.substr(0, s.size() - match[sz - 1]) + s; 23 } 24 };
上边程序可以解释如下:
对构造的字符串combine进行匹配操作后,得到如下结果:
匹配部分也就是s和rev_s重复的部分,而不匹配的部分就是它们不一样的部分。接下来的字符串拼接操作就是:
拼接完成后即是最终的结果。