LeetCode Online Judge 题目C# 练习 - Scramble String

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.
Below is one possible representation of s1 = "great":
 great
　 / \
 gr eat
 / \ / \
g r e at
　　  / \
       a t
To scramble the string, we may choose any non-leaf node and swap its two children.
For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".
 rgeat
   / \
 rg eat
 / \ / \
r g e at
       / \
       a t
We say that "rgeat" is a scrambled string of "great".
Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".
 rgtae
   / \
 rg tae
 / \ / \
r g ta e
     / \
     t a
We say that "rgtae" is a scrambled string of "great".
Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

 1         public static bool ScrambleString(string s1, string s2)

 2         {

 3             if (s1 == s2)

 4                 return true;

 5             else if (s1.Length != s2.Length)

 6                 return false;

 7             else if (s1.Length == 1 && s1 != s2)

 8                 return false;

 9             else

10             {

11                 for (int i = 1; i < s1.Length; i++)

12                 {

13                     if ((ScrambleString(s1.Substring(0, i), s2.Substring(0, i)) && ScrambleString(s1.Substring(i), s2.Substring(i))) ||

14                         ScrambleString(s1.Substring(0, i), s2.Substring(s2.Length - i)) && ScrambleString(s1.Substring(i), s2.Substring(0, s2.Length - i)))

15                     {

16                         return true;

17                     }

18                 }

19                 return false;

20             }

21         }

代码分析：

　　递归，看来有关树的题，用递归还是八九不离十了。

　　这段code过不了在LeetCdoe的Large Case。改成下面这样才可以。 其实时间复杂度一样

　　

 1         public static bool ScrambleStringOpt(string s1, string s2)

 2         {

 3             if (s1 == s2)

 4                 return true;

 5             else if (s1.Length != s2.Length)

 6                 return false;

 7             else if (s1.Length == 1 && s1 != s2)

 8                 return false;

 9             else

10             {

11                 int n = s1.Length;

12                 for (int i = 1; i <= n / 2; i++)

13                 {

14                     string a1 = s1.Substring(0, i), b1 = s1.Substring(i), a2 = s2.Substring(0, i), b2 = s2.Substring(i);

15                     string c1 = s1.Substring(0, n - i), d1 = s1.Substring(n - i), c2 = s2.Substring(0, n - i), d2 = s2.Substring(n - i);

16 

17                     if (ScrambleStringOpt(a1, a2) && ScrambleStringOpt(b1, b2))

18                         return true;

19                     if (ScrambleStringOpt(a1, d2) && ScrambleStringOpt(b1, c2))

20                         return true;

21                     if (ScrambleStringOpt(d1, a2) && ScrambleString(c1, b2))

22                         return true;

23                     if (ScrambleStringOpt(d1, d2) && ScrambleString(c1, c2))

24                         return true;

25                 }

26                 return false;

27             }

28         }

代码分析：

　　要提升效率， && 操作符，把做得快的放在前面！如果把做的慢的放前面的话，LeetCode Large Case 过不了。。。