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 = 0;
        int subEnd = 0;
        int rowIndex,columnIndex;
        
        for(rowIndex = 0; rowIndex < sLen; ++ rowIndex)
        {
            charRelFlag[rowIndex][rowIndex] = true;
            for(columnIndex = rowIndex + 1; columnIndex < sLen; ++ columnIndex)
            {
                charRelFlag[rowIndex][columnIndex] = false;
            }
        }
        
        int maxLen = 1;
        for(columnIndex = 1; columnIndex < sLen; ++ columnIndex)
        {
            for(rowIndex = 0; rowIndex < columnIndex; ++ rowIndex)
            {
                if(s[rowIndex] == s[columnIndex])
                {
                    if(rowIndex + 1 == columnIndex)
                    {
                        charRelFlag[rowIndex][columnIndex] = true;
                        if(maxLen < 2)
                        {
                            maxLen = 2;
                            subStart = rowIndex;
                            subEnd = columnIndex;
                        }
                    }else
                    {
                        if(charRelFlag[rowIndex + 1][columnIndex - 1] == true)
                        {
                            charRelFlag[rowIndex][columnIndex] = true;
                            if(maxLen < columnIndex - rowIndex + 1)
                            {
                                maxLen = columnIndex - rowIndex + 1;
                                subStart = rowIndex;
                                subEnd = columnIndex;
                            }
                        }
                    }
                    
                }
            }
        }
        
        return s.substr(subStart,maxLen);
    }
};