题目描述
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”:
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great
/
gr eat
/ /
g r e at
/
a t
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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”.
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rgeat
/
rg eat
/ /
r g e at
/
a t
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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”.
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rgtae
/
rg tae
/ /
r g ta e
/
t a
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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.
代码
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public class {
public boolean isScramble(String s1, String s2) {
int len1 = s1.length(), len2 = s2.length();
if(len1 != len2) return false;
if(len1 == 0) return true;
if(s1.equals(s2)) return true;
int[] letters = new int[256];
for(int i = 0; i < len1; i++){
letters[s1.charAt(i)]++;
letters[s2.charAt(i)]--;
}
for(int i = 0; i < len1; i++){
if(letters[s1.charAt(i)] != 0) return false;
}
for(int i = 1; i < len1; i++){
if(isScramble(s1.substring(0, i), s2.substring(0,i)) && isScramble(s1.substring(i), s2.substring(i))) return true;
if(isScramble(s1.substring(0, i), s2.substring(len2 - i)) && isScramble(s1.substring(i), s2.substring(0,len2-i))) return true;
}
return false;
}
}
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