Leetcode 647: Palindromic Substrings

最新推荐文章于 2021-08-08 18:28:40 发布

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最新推荐文章于 2021-08-08 18:28:40 发布

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原文链接：http://www.cnblogs.com/liangmou/p/8384493.html

本文介绍了一种算法，用于计算给定字符串中所有不同回文子串的数量。通过两个不同的实现方式，展示了如何利用动态规划及哈希表来解决这一问题，并提供了详细的代码解析。

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Input: "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".

Example 2:

Input: "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".

Note:

The input string length won't exceed 1000.

 1 public class Solution {
 2     public int CountSubstrings(string s) {
 3         if (s == null || s.Length == 0) return 0;
 4         
 5         var dp = new bool[s.Length, s.Length];
 6         
 7         int count = 0;
 8         
 9         for (int i = s.Length - 1; i >= 0; i--)
10         {
11             for (int j = i; j < s.Length; j++)
12             {
13                 if ((s[i] == s[j]) && (j - i <= 2  || dp[i + 1, j - 1]))
14                 {
15                     dp[i, j] = true;
16                     count++;
17                 }
18             }
19         }
20         
21         return count;
22     }
23 }
24 
25 public class Solution1 {
26     public int CountSubstrings(string s) {
27         if (s == null || s.Length == 0) return 0;
28         
29         int count = 0;
30         var dict = new Dictionary<Tuple<int, int>, bool>();
31         
32         for (int i = s.Length - 1; i >= 0; i--)
33         {
34             for (int j = i; j < s.Length; j++)
35             {
36                 if (IsPalindrome(s, i, j, dict)) count++;    
37             }
38         }
39         
40         return count;
41     }
42     
43     private bool IsPalindrome(string s, int i, int j, Dictionary<Tuple<int, int>, bool> dict)
44     {
45         var key = new Tuple<int, int>(i, j);
46         
47         if (i == j)
48         {
49             dict[key] = true;
50             return true;
51         }
52         
53         if (i + 1 == j)
54         {
55             if (s[i] == s[j])
56             {
57                 dict[key] = true;
58                 return true;
59             }
60             else
61             {
62                 dict[key] = false;
63                 return false;
64             }
65         }
66         
67         var key1 = new Tuple<int, int>(i + 1, j - 1);
68         
69         // since we have aleady processed them before, we don't need to process again
70         if (s[i] == s[j] && dict.ContainsKey(key1) && dict[key1])
71         {
72             dict[key] = true;
73             return true;
74         }
75         else
76         {
77             dict[key] = false;
78             return false;
79         }
80     }
81 }