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Scramble String leetcode java
题目:
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.
题解:
这道题完全转载code ganker(http://blog.csdn.net/linhuanmars/article/details/24506703):
“这道题看起来是比较复杂的,如果用brute force,每次做切割,然后递归求解,是一个非多项式的复杂度,一般来说这不是面试官想要的答案。
这其实是一道三维动态规划的题目,我们提出维护量res[i][j][n],其中i是s1的起始字符,j是s2的起始字符,而n是当前的字符串长度,res[i][j][len]表示的是以i和j分别为s1和s2起点的长度为len的字符串是不是互为scramble。
有了维护量我们接下来看看递推式,也就是怎么根据历史信息来得到res[i][j][len]。判断这个是不是满足,其实我们首先是把当前s1[i...i+len-1]字符串劈一刀分成两部分,然后分两种情况:第一种是左边和s2[j...j+len-1]左边部分是不是scramble,以及右边和s2[j...j+len-1]右边部分是不是scramble;第二种情况是左边和s2[j...j+len-1]右边部分是不是scramble,以及右边和s2[j...j+len-1]左边部分是不是scramble。如果以上两种情况有一种成立,说明s1[i...i+len-1]和s2[j...j+len-1]是scramble的。而对于判断这些左右部分是不是scramble我们是有历史信息的,因为长度小于n的所有情况我们都在前面求解过了(也就是长度是最外层循环)。
上面说的是劈一刀的情况,对于s1[i...i+len-1]我们有len-1种劈法,在这些劈法中只要有一种成立,那么两个串就是scramble的。
总结起来递推式是res[i][j][len] = || (res[i][j][k]&&res[i+k][j+k][len-k] || res[i][j+len-k][k]&&res[i+k][j][len-k]) 对于所有1<=k<len,也就是对于所有len-1种劈法的结果求或运算。因为信息都是计算过的,对于每种劈法只需要常量操作即可完成,因此求解递推式是需要O(len)(因为len-1种劈法)。
如此总时间复杂度因为是三维动态规划,需要三层循环,加上每一步需要线行时间求解递推式,所以是O(n^4)。虽然已经比较高了,但是至少不是指数量级的,动态规划还是有很大有事的,空间复杂度是O(n^3)。代码如下:”
代码如下:
2 if(s1==null || s2==null || s1.length()!=s2.length())
3 return false;
4 if(s1.length()==0)
5 return true;
6 boolean[][][] res = new boolean[s1.length()][s2.length()][s1.length()+1];
7 for(int i=0;i<s1.length();i++)
8 {
9 for(int j=0;j<s2.length();j++)
10 {
11 res[i][j][1] = s1.charAt(i)==s2.charAt(j);
12 }
13 }
14 for(int len=2;len<=s1.length();len++)
15 {
16 for(int i=0;i<s1.length()-len+1;i++)
17 {
18 for(int j=0;j<s2.length()-len+1;j++)
19 {
20 for(int k=1;k<len;k++)
21 {
22 res[i][j][len] |= res[i][j][k]&&res[i+k][j+k][len-k] || res[i][j+len-k][k]&&res[i+k][j][len-k];
23 }
24 }
25 }
26 }
27 return res[0][0][s1.length()];
28 }
同样这道题也可以用递归来做。
我个人觉得用递归做更加方便简单些,思路还是上面那个思路,分两刀那种(底下解释引用自:http://blog.unieagle.net/2012/10/23/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Ascramble-string%EF%BC%8C%E4%B8%89%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/):
“简单的说,就是s1和s2是scramble的话,那么必然存在一个在s1上的长度l1,将s1分成s11和s12两段,同样有s21和s22。
那么要么s11和s21是scramble的并且s12和s22是scramble的;要么s11和s22是scramble的并且s12和s21是scramble的。”
代码:
2 if(s1.length() != s2.length())
3 return false;
4
5 if(s1.length()==1 && s2.length()==1)
6 return s1.charAt(0) == s2.charAt(0);
7
8 char[] t1 = s1.toCharArray(), t2 = s2.toCharArray();
9 Arrays.sort(t1);
10 Arrays.sort(t2);
11 if(!new String(t1).equals(new String(t2)))
12 return false;
13
14 if(s1.equals(s2))
15 return true;
16
17 for(int split = 1; split < s1.length(); split++){
18 String s11 = s1.substring(0, split);
19 String s12 = s1.substring(split);
20
21 String s21 = s2.substring(0, split);
22 String s22 = s2.substring(split);
23 if(isScramble(s11, s21) && isScramble(s12, s22))
24 return true;
25
26 s21 = s2.substring(0, s2.length() - split);
27 s22 = s2.substring(s2.length() - split);
28 if(isScramble(s11, s22) && isScramble(s12, s21))
29 return true;
30 }
31 return false;
32 }
Reference:
http://www.cnblogs.com/lichen782/p/leetcode_Scramble_String.html
http://blog.unieagle.net/2012/10/23/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Ascramble-string%EF%BC%8C%E4%B8%89%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/
http://blog.csdn.net/fightforyourdream/article/details/17707187
http://blog.csdn.net/linhuanmars/article/details/24506703