Scramble String
Desicription
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
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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"
.
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"
.
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
.
Solution
class Solution {
public:
bool isScramble(string s1, string s2) {
if(s1 == s2)
return 1;
vector<int> cnt(26, 0);
for(int i = 0; s1[i]; i++)
cnt[s1[i] - 'a']++, cnt[s2[i] - 'a']--;
for(int i = 0; i < 26; i++)
if(cnt[i])
return 0;
for(int i = 1; s1[i]; i++){
if(isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)))
return 1;
if(isScramble(s1.substr(0, i), s2.substr(s1.size() - i)) && isScramble(s1.substr(i), s2.substr(0, s1.size() - i)))
return 1;
}
return 0;
}
};
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