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.

链接: http://leetcode.com/problems/scramble-string/

题解:

题目比较长,理解起来也很费力。判断两个string是否互为scramble。卡了很久没有思路,今天早上坐PATH的时候看到一些讲解觉得很不错,下午试了试觉得可以。下面是用DFS + 剪枝。

Time Complexity - O(4n), Space Complexity - O(n)。

public class Solution {
public boolean isScramble(String s1, String s2) {
if(s1 == null || s2 == null || s1.length() != s2.length())
return false;
if(s1.equals(s2))
return true;
char[] arr1 = s1.toCharArray();
char[] arr2 = s2.toCharArray();
Arrays.sort(arr1);
Arrays.sort(arr2);
if(!new String(arr1).equals(new String(arr2)))
return false; for(int i = 1; i < s1.length(); i++) {
String s11 = s1.substring(0, i);
String s12 = s1.substring(i);
String s21 = s2.substring(0, i);
String s22 = s2.substring(i);
if(isScramble(s11, s21) && isScramble(s12, s22)) //left - left , right - right
return true;
s21 = s2.substring(0, s2.length() - i);
s22 = s2.substring(s2.length() - i);
if(isScramble(s11, s22) && isScramble(s12, s21)) //left - right, right - left
return true;
} return false;
}
}

还有一种做法是三维DP,还要仔细研究一下。

二刷:

还是recursive比较好理解一些,三维dp以后再说了。下面分析一下recursive的几个点:

  1. 首先判断边界
  2. 其次,当s1等于s2的时候,我们判断可以返回如true
  3. 否则我们对排序后的 s1和s2进行一个比较,假如不等,则我们舍去
  4. 否则我们进入遍历的循环体,注意starting index是从1开始
    1. 我们设置s11, s12, s21和s22,然后递归判断(s11, s21)以及(s12和s22)这两个pair,假如同时满足scramble,则我们可以返回true
    2. 否则,我们尝试交换过一次的结果,即重设s21和s22从尾部开始split。然后比较新的(s11, s22)以及(s12和s21)这两个pair,假如同时满足条件则返回true
  5. 否则返回false
  6. 复杂度的来说 ,不考虑substring的复杂度话, recursive depth大约是n,branching factor是4,所以时间复杂度是大约是O(n4), 空间复杂度也是O(n4)

Java:

Time Complexity - O(4n), Space Complexity - O(n4)。  这里可能算得还是不对,希望有机会能再算算。

public class Solution {
public boolean isScramble(String s1, String s2) {
if (s1 == null || s2 == null || s1.length() != s2.length()) {
return false;
}
if (s1.equals(s2)) {
return true;
}
char[] arr1 = s1.toCharArray();
char[] arr2 = s2.toCharArray();
Arrays.sort(arr1);
Arrays.sort(arr2);
if (!String.valueOf(arr1).equals(String.valueOf(arr2))) {
return false;
}
int len = s1.length();
for (int i = 1; i < len; i++) {
String s11 = s1.substring(0, i);
String s12 = s1.substring(i);
String s21 = s2.substring(0, i);
String s22 = s2.substring(i);
if (isScramble(s11, s21) && isScramble(s12, s22)) {
return true;
}
s21 = s2.substring(0, len - i);
s22 = s2.substring(len - i);
if (isScramble(s11, s22) && isScramble(s12, s21)) {
return true;
}
}
return false;
}
}

题外话:

2/9/2016

二刷进度一直比较慢,今天要开始加快速度了,准备开启糙快猛节奏。

Reference:

https://leetcode.com/discuss/46803/accepted-java-solution

https://leetcode.com/discuss/36470/share-my-4ms-c-recursive-solution

https://leetcode.com/discuss/3632/any-better-solution

https://leetcode.com/discuss/2504/can-you-partition-string-index-any-time-producing-scramble

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

http://www.cnblogs.com/jianxinzhou/p/4712148.html

https://www.slyar.com/blog/depth-first-search-even-odd-pruning.html

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