LeetCode:5_Longest Palindromic Substring | 最长的回文子串 | Medium

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LeetCode:5_Longest Palindromic Substring | 最长的回文子串 | Medium

chambai 2016-04-06 10:25:00 浏览752
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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.

解题思路:
1、简单思路:暴力破解法,时间复杂度O(n^3),肯定通不过。

2、动态规划法:(一般含“最XX”等优化词义的题意味着都可以动态规划求解),时间复杂度O(n^2),空间复杂度O(n^2)。

形如"abba", "abbba"这样的字符串,如果用dp[i][j]表示从下标i到j之间的子字符串所构成的回文子串的长度,则有:

dp[i][j] = dp[i+1][j-1] && s[i] == s[j]

初始化:

奇数个字符:dp[i][i] = 1; 偶数个字符:dp[i][i+1] = 1

 1 //动态规划法
 2 string LongestPalindromicStringDP(string str)
 3 {
 4     size_t n = str.size();
 5     if (n == 0 || n == 1)
 6         return str;
 7     //创建dp二维数组
 8     //d[i][j] = dp[i+1][j-1] && s[i] == s[j]
 9     bool **dp = new bool *[n];
10     for (size_t i = 0; i < n; ++ i)
11         dp[i] = new bool[n];
12     
13     int nLongestIndex = 0; //最长回文子串的开始下标
14     int nMaxLen = 0; //长度
15 
16     for (size_t i = 0; i < n; i ++)
17         for (size_t j = 0; j < n; j ++)
18             dp[i][j] = false;
19     for (size_t i = 0; i < n; ++ i) 
20         dp[i][i] = true; //初始化一个字母
21 
22     for (size_t i = 0; i < n - 1; ++ i) {
23         if (str[i] == str[i+1]) {
24             dp[i][i+1] = true; //初始化两个相同的字母"aa"
25             nLongestIndex = i;
26             nMaxLen = 2;
27         }
28     }
29     //从长度3开始操作, (aba)ba, a(bab)a, ab(aba)
30     for (size_t len = 3; len <= n; ++ len) {
31         for (size_t i = 0; i < n-len+1; ++ i) {
32             size_t j = i + len - 1;
33             if (str[i] == str[j] && dp[i+1][j-1]){
34                 dp[i][j] = true;
35                 nLongestIndex = i;
36                 nMaxLen = len;
37             }
38         }
39     }
40 
41     //释放dp
42     for (size_t i = 0; i < n; ++ i)
43         delete []dp[i];
44     delete []dp;
45     
46     return str.substr(nLongestIndex, nMaxLen);
47 }

 

3、中心扩散法:时间复杂度O(n^2),空间复杂度O(1)
顾名思义,从一个节点,分别向两边扩散,用两个指针分别向前向后遍历。

 1 //中心扩散法
 2 string LongestPalindromicString(string str) 
 3 {
 4     size_t n = str.size();
 5     if (n == 0 || n == 1) 
 6         return str;
 7 
 8     size_t start = 0;
 9     size_t nMaxLen = 0;
10 
11     for (size_t i = 0; i < n; ++ i) {
12         size_t j = i, k = i; //分别从中心向两边扩散
13         while(k < n-1 && str[k] == str[k+1])
14             k++; //有相同字母的情况:"aaa"
15         while(j > 0 && k < n-1 && str[j-1] == str[k+1]){
16             k++; //不同字母情况: "aba"
17             j--;
18         }
19         if(k-j+1 > nMaxLen){
20             nMaxLen = k-j+1;
21             start = j;
22         }
23     }
24     return str.substr(start, nMaxLen);
25 }

 

4、很明显,这两种方法时间复杂度还是太高了,还有一种算法叫Manacher,时间复杂度能够降为O(n),空间复杂度也为O(n),先记下,以后再做研究吧。

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