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algorithm/PuzzleCoding/src/ScrambleString.java at master · wodgithub/algorithm · GitHub
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PuzzleCoding
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src
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ScrambleString.java
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algorithm
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PuzzleCoding
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src
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ScrambleString.java
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/**
* Scramble string, for two strings, say str1 = “tiger” and str2 = “itreg”
* <p/>
* Scramble string, for two strings, say str1 = “tiger” and str2 = “itreg”
* tiger
* / \
* ti ger
* / \ / \
* t i g er
* / \
* e r
* <p/>
* itreg
* / \
* it reg
* / \ / \
* t i g re
* / \
* e r
* <p/>
* In above case, since s1 can be generated from str2, that is, str1 can be generated from the partition tree,
* which has the same leaf of str2, but the order of its child can be switched.
* So str1 is said to be a scramble string of str2.
* <p/>
* A counter example is that Tiger and Tgrie are not scramble string of each other
* <p/>
* The principle of this approach goes as following
* 1) Given two strings A and B, try to split each of them into two sub-strings sA1,sA2,sB1,
* sB2 where sA1 and sB1 are anagrams and sA2 and sB2 are anagrams.
* 2) If sA1 and sB1 are the scramble strings while sA2 and sBs are scramble strings,
* then A and B should be scramble strings.
* 3) Recursively analyze sA1 and sB1, sA2 and sB2 by start over from step 1).
* <p/>
* In previous example
* <p/>
* str1 = aabbbaccd, str2: aabcdbcba,
* <p/>
* They are split as: (aab)(bbaccd), (aab)(cdbcba) => compare bbaccd and cdbcba
* They are split as: (bbac)(cd), (cd)(bcba) => compare bbac and bcba
* They are split as: (b)(bac), b(cba) => (ba)(c) vs (c)(ba)
*/
public
class
ScrambleString
{
public
static
void
main
(
String
[]
args
) {
// they are anagram as a whole string, but not scramble
String
s1
=
"tiger"
;
String
s2
=
"tgrie"
;
System
.
out
.
println
(
isScrambleStringRecursive
(
s1
,
s2
));
// they are anagram as well as scramble
s1
=
"tiger"
;
s2
=
"itreg"
;
System
.
out
.
println
(
isScrambleStringRecursive
(
s1
,
s2
));
}
public
static
boolean
isScrambleStringRecursive
(
String
s1
,
String
s2
) {
if
(
s1
.
length
() !=
s2
.
length
())
return
false
;
if
(
s1
.
equals
(
s2
))
return
true
;
else
{
for
(
int
i
=
0
;
i
<
s1
.
length
() -
1
;
i
++) {
String
s1a
=
s1
.
substring
(
0
,
i
+
1
);
String
s1b
=
s1
.
substring
(
i
+
1
);
String
s2a
=
s2
.
substring
(
0
,
i
+
1
);
String
s2b
=
s2
.
substring
(
i
+
1
);
String
s2ar
=
s2
.
substring
(
s2
.
length
() -
1
-
i
);
String
s2br
=
s2
.
substring
(
0
,
s2
.
length
() -
1
-
i
);
if
((
isScrambleStringRecursive
(
s1a
,
s2a
) &&
isScrambleStringRecursive
(
s1b
,
s2b
)) ||
(
isScrambleStringRecursive
(
s1a
,
s2ar
) &&
isScrambleStringRecursive
(
s1b
,
s2br
))) {
return
true
;
}
}
return
false
;
}
}
}
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