Shirlies
宁静专注认真的程序媛~

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

 1 #define MAX_LENGTH 1001
 2 class Solution {
 3 public:
 4     string longestPalindrome(string s) {
 5         if(s.empty())
 6         {
 7             return string("");
 8         }
 9         bool charRelFlag[MAX_LENGTH][MAX_LENGTH];
10         int sLen = s.size();
11         int subStart = 0;
12         int subEnd = 0;
13         int rowIndex,columnIndex;
14         
15         for(rowIndex = 0; rowIndex < sLen; ++ rowIndex)
16         {
17             charRelFlag[rowIndex][rowIndex] = true;
18             for(columnIndex = rowIndex + 1; columnIndex < sLen; ++ columnIndex)
19             {
20                 charRelFlag[rowIndex][columnIndex] = false;
21             }
22         }
23         
24         int maxLen = 1;
25         for(columnIndex = 1; columnIndex < sLen; ++ columnIndex)
26         {
27             for(rowIndex = 0; rowIndex < columnIndex; ++ rowIndex)
28             {
29                 if(s[rowIndex] == s[columnIndex])
30                 {
31                     if(rowIndex + 1 == columnIndex)
32                     {
33                         charRelFlag[rowIndex][columnIndex] = true;
34                         if(maxLen < 2)
35                         {
36                             maxLen = 2;
37                             subStart = rowIndex;
38                             subEnd = columnIndex;
39                         }
40                     }else
41                     {
42                         if(charRelFlag[rowIndex + 1][columnIndex - 1] == true)
43                         {
44                             charRelFlag[rowIndex][columnIndex] = true;
45                             if(maxLen < columnIndex - rowIndex + 1)
46                             {
47                                 maxLen = columnIndex - rowIndex + 1;
48                                 subStart = rowIndex;
49                                 subEnd = columnIndex;
50                             }
51                         }
52                     }
53                     
54                 }
55             }
56         }
57         
58         return s.substr(subStart,maxLen);
59     }
60 };
View Code

 

posted on 2016-08-08 15:03  Shirlies  阅读(271)  评论(0编辑  收藏  举报