LeetCode学习（5）Shortest Palindrome（最短回文串）

1.问题描述

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”.

给定一个字符串，在字符串的前面加入字符，将该字符串变成一个最短的回文字符串，如题目中所示aacecaaa变成最短的回文只需要在前面加上字符a ，得到回文字符串aaacecaaa

2.代码时刻

//c++ VS2013
#include<iostream>
#include<string>
using namespace std;
bool Check(string s, int low, int high)
{
//该函数主要判断从low到high的字符串是否为回文字符串
if (low == high)
return true;
while (low<high)
{
if (s[low] != s[high])
return false;
low++;
high--;
}
return true;
}
string ShortestPalindrome(string s)
{
int i, j;
int len;
string result = "";
len = s.length() - 1;
if (len <= 0)
return "";
for (len; len>0; len--)  //从输入的字符串最后一位开始往前找，找到满足回文的最大回文字符串
{
if (s[0] == s[len] && Check(s, 0, len))
break;
}
/*找到第1-len位表示最长的回文串，第(len+1)-length()位就是没有匹配上的，反转第(len+1)-length()位并加在原来字符串最前面就是*/
for (i = s.length() - 1; i>len; i--)
result += s[i];
result += s;
return result;
}
int main()
{
string s = "xixl";
string a = ShortestPalindrome(s);
cout<<a << endl;
return 0;
}

3.测试结果

测试用例： s = “xixl”

测试结果： s=”lxixil”

测试用例： s = “aaacaaafigh”

测试结果： s = “hgifaaacaaafigh”