Leetcode---Longest Palindromic Substring

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.

首先我用了DP，没想到存储没报错，时间报错了，TLE了。

原理是记录所有回文，然后更新最大值。

比较值得注意的是

1. for(int i=n-2;i>=0;i--){
   
2.         for(int j=i+1;j<=n-1;j++){
   
3.             dp[i][j]=dp[i+1][j-1]&&(s[i]==s[j]);

i依赖于i+1, j依赖于j-1，所以i--，j++的遍历

1. string longestPalindrome(string s) {
   
2.     int n=s.length();
   
3.     vector<vector<bool>> dp(n,vector<bool>(n,false));
   
4.     for(int i=0;i<n;i++){
   
5.         for(int j=0;j<=i;j++){
   
6.             dp[i][j]=true;
   
7.         }
   
8.     }
   
9.     int num=1;
   
10.     int x=0;
    
11.     int y=0;
    
12.     for(int i=n-2;i>=0;i--){
    
13.         for(int j=i+1;j<=n-1;j++){
    
14.             dp[i][j]=dp[i+1][j-1]&&(s[i]==s[j]);
    
15.             if(dp[i][j] && (j-i+1>num)){
    
16.                 num=j-i+1;
    
17.                 x=i;
    
18.                 y=j;
    
19.             }
    
20.         }
    
21.     }
    
22.     return s.substr(x,num);
    
23. }

后来发现其实朴素的判断回文空间复杂性更小。其实就是从0到n-1遍历，每个节点为中心判断回文，分为奇偶两种情况，

奇数遍历：

1. for (int i = 1; i < s.length(); ++i){
   
2.         low = i - 1;
   
3.         high = i + 1;
   
4.         while (low >= 0 && high <s.length() && s[low] == s[high])
   
5.         {
   
6.             //是回文
   
7.             --low;
   
8.             ++high;
   
9.         }
   
10. }

偶 数遍历：

1. for (int i = 1; i < s.length(); ++i){
   
2.         low = i - 1;
   
3.         high = i;
   
4.         while (low >= 0 && high <s.length() && s[low] == s[high])
   
5.         {
   
6.             //是回文
   
7.             --low;
   
8.             ++high;
   
9.         }
   
10. }

代码如下：

1. string longestPalindrome(string s) {
   
2. 
   
3. int start = 0;
   
4.  int maxLength = 1;
   
5. string res;
   
6. 
   
7.     if(s.length()==1)
   
8.        return s;
   
9.     int low, high;
   
10.     for (int i = 1; i < s.length(); ++i)
    
11.     {
    
12.         low = i - 1;
    
13.         high = i;
    
14.         while (low >= 0 && high < s.length() && s[low] == s[high])
    
15.         {
    
16.             if (high - low + 1 > maxLength)
    
17.             {
    
18.                 start = low;
    
19.                 maxLength = high - low + 1;
    
20.             }
    
21.             --low;
    
22.             ++high;
    
23.         }
    
24.          low = i - 1;
    
25.         high = i + 1;
    
26.         while (low >= 0 && high <s.length() && s[low] == s[high])
    
27.         {
    
28.             if (high - low + 1 > maxLength)
    
29.             {
    
30.                 start = low;
    
31.                 maxLength = high - low + 1;
    
32.             }
    
33.             --low;
    
34.             ++high;
    
35.         }
    
36.     }
    
37.      res = s.substr(start,maxLength);
    
38. return res;
    
39. 
    
40. 
    
41. }

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