leetcode-【中等题】5. Longest Palindromic Substring

题目

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

答案

一道动规题目，每次用当前位置columnIndex跟前面位置rowIndex去比较时，如果相同的话，就去看columnIndex- 1和rowIndex+1字母的状态，其实在走到columnIndex时，columnIndex-1和它自己前面的字母是比较过了的，将这个状态记下来就行。

讲真，我这代码效率不高，应该还有其他思路，再想想。

代码

 #define MAX_LENGTH 1001
 class Solution {
 public:
     string longestPalindrome(string s) {
         if(s.empty())
         {
             return string("");
         }
         bool charRelFlag[MAX_LENGTH][MAX_LENGTH];
         int sLen = s.size();
         int subStart = ;
         int subEnd = ;
         int rowIndex,columnIndex;

         for(rowIndex = ; rowIndex < sLen; ++ rowIndex)
         {
             charRelFlag[rowIndex][rowIndex] = true;
             for(columnIndex = rowIndex + ; columnIndex < sLen; ++ columnIndex)
             {
                 charRelFlag[rowIndex][columnIndex] = false;
             }
         }

         int maxLen = ;
         for(columnIndex = ; columnIndex < sLen; ++ columnIndex)
         {
             for(rowIndex = ; rowIndex < columnIndex; ++ rowIndex)
             {
                 if(s[rowIndex] == s[columnIndex])
                 {
                     if(rowIndex +  == columnIndex)
                     {
                         charRelFlag[rowIndex][columnIndex] = true;
                         if(maxLen < )
                         {
                             maxLen = ;
                             subStart = rowIndex;
                             subEnd = columnIndex;
                         }
                     }else
                     {
                         if(charRelFlag[rowIndex + ][columnIndex - ] == true)
                         {
                             charRelFlag[rowIndex][columnIndex] = true;
                             if(maxLen < columnIndex - rowIndex + )
                             {
                                 maxLen = columnIndex - rowIndex + ;
                                 subStart = rowIndex;
                                 subEnd = columnIndex;
                             }
                         }
                     }

                 }
             }
         }

         return s.substr(subStart,maxLen);
     }
 };