Google+ Followers

Monday, December 24, 2012

[LeetCode] Longest Palindromic Substring 解题报告


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.
» Solve this problem

[解题思路]
O(n*n)。对于每一个字符,以之作为中间元素往左右寻找。注意处理奇偶两种模式:
1. aba
2. abba

1:  string longestPalindrome(string s) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      int startIndex = 0;  
5:      int len = 0;            
6:      int sI,eI;  
7:      for(int i =0; i< s.size() -1; i++)  
8:      {  
9:           if(s[i] == s[i+1])  
10:          {  
11:               sI = i;  
12:               eI = i+1;  
13:               Search(s, sI, eI, len, startIndex);  
14:          }  
15:          sI = i;  
16:          eI = i;  
17:          Search(s, sI, eI, len, startIndex);  
18:      }       
19:      if(len == 0)  
20:        len = s.size();  
21:      return s.substr(startIndex, len);  
22:    }  
23:    void Search(string&s, int sI, int eI, int&len, int& startIndex)  
24:    {  
25:      int step = 1;   
26:      while((sI-step)>=0&& (eI+step)<s.size())  
27:      {  
28:        if(s[sI-step] != s[eI+step])  
29:        {       
30:          break;  
31:        }                      
32:        step++;  
33:      }  
34:      int wid = eI- sI+2*step -1;  
35:      if(wid > len)  
36:      {  
37:        len = wid;  
38:        startIndex = sI - step+1;   
39:      }  
40:    }  

此题也有O(n)解法。相当巧妙,http://www.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html


Version2: add DP solution, 3/5/2013
定义函数
P[i,j] = 字符串区间[i,j]是否为palindrome.

首先找个例子,比如S="abccb",
    S=    a  b  c  c  b
Index = 0  1  2  3  4

P[0,0] =1  //each char is a palindrome
P[0,1] =S[0] == S[1]    , P[1,1] =1
P[0,2] = S[0] == S[2] && P[1,1], P[1,2] = S[1] == S[2] , P[2,2] = 1
P[0,3] = S[0] == S[3] && P[1,2], P[1,3] = S[1] == S[3] && P[2,2] , P[2,3] =S[2] ==S[3],  P[3,3]=1      
......................
由此就可以推导出规律

P[i,j] = 1  if i ==j
        =  S[i] ==S[j]   if j = i+1
        =  S[i] == S[j] && P[i+1][j-1]  if j>i+1

实现如下:
1:       string longestPalindrome(string s) {  
2:            int len = s.size();  
3:            int P[len][len];  
4:            memset(P, 0, len*len*sizeof(int));  
5:            int maxL=0, start=0, end=0;  
6:            for(int i =0; i< s.size(); i++)  
7:            {  
8:                 for(int j =0; j<i; j++)  
9:                 {  
10:                      P[j][i] = (s[j] == s[i] && (i-j<2 || P[j+1][i-1]));  
11:                      if(P[j][i] && maxL < (i-j+1))  
12:                      {  
13:                           maxL = i-j+1;  
14:                           start = j;  
15:                           end = i;  
16:                      }  
17:                 }  
18:                 P[i][i] =1;  
19:            }  
20:            return s.substr(start, end-start +1);  
21:       }  






Post a Comment