FazBrowse GitHub Viewer
|
Trending
|
URL:
|
Home
Tools:
[Download Repo ZIP]
[View Raw Code]
[Original HTTPS Page]
leetcode-1/src/scrambleString/scrambleString.cpp at master · daction/leetcode-1 · GitHub
daction
/
leetcode-1
Public
forked from
haoel/leetcode
Notifications
You must be signed in to change notification settings
Fork
0
Star
0
Code
Pull requests
0
Actions
Projects
Wiki
Security and quality
0
Insights
Additional navigation options
Code
Pull requests
Actions
Projects
Wiki
Security and quality
Insights
Expand file tree
Breadcrumbs
leetcode-1
/
src
/
scrambleString
/
scrambleString.cpp
Copy path
More file actions
More file actions
Latest commit
History
History
History
180 lines (162 loc) · 4.64 KB
Breadcrumbs
leetcode-1
/
src
/
scrambleString
/
scrambleString.cpp
Copy path
File metadata and controls
180 lines (162 loc) · 4.64 KB
Raw
Copy raw file
Download raw file
Open symbols panel
Edit and raw actions
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
//
Source : https://oj.leetcode.com/problems/scramble-string/
//
Author : Hao Chen
//
Date : 2014-10-09
/*
*********************************************************************************
*
* 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.
*
*
*********************************************************************************
*/
#
include
<
stdlib.h
>
#
include
<
time.h
>
#
include
<
iostream
>
#
include
<
vector
>
#
include
<
string
>
#
include
<
algorithm
>
using
namespace
std
;
//
The recursive way is quite simple.
//
1) break the string to two parts:
//
s1[0..j] s1[j+1..n]
//
s2[0..j] s2[j+1..n]
//
2) then
//
isScramble(s1[0..j], s2[0..j]) && isScramble(s1[j+1..n], s2[j+1..n])
//
OR
//
isScramble(s1[0..j], s2[j+1, n]) && isScramble(s1[j+1..n], s2[0..j])
bool
isScramble_recursion
(string s1, string s2) {
if
(s1.
size
()!= s2.
size
() || s1.
size
()==
0
|| s2.
size
()==
0
) {
return
false
;
}
if
(s1 == s2){
return
true
;
}
string ss1 = s1;
string ss2 = s2;
sort
(ss1.
begin
(), ss1.
end
());
sort
(ss2.
begin
(), ss2.
end
());
if
(ss1 != ss2 ) {
return
false
;
}
for
(
int
i=
1
; i<s1.
size
(); i++) {
if
(
isScramble_recursion
(s1.
substr
(
0
,i), s2.
substr
(
0
,i)) &&
isScramble_recursion
(s1.
substr
(i, s1.
size
()-i), s2.
substr
(i, s2.
size
()-i)) ) {
return
true
;
}
if
(
isScramble_recursion
(s1.
substr
(
0
,i), s2.
substr
(s2.
size
()-i, i)) &&
isScramble_recursion
(s1.
substr
(i, s1.
size
()-i), s2.
substr
(
0
, s2.
size
()-i)) ) {
return
true
;
}
}
return
false
;
}
/*
* Definition
*
* dp[k][i][j] means:
*
* a) s1[i] start from 'i'
* b) s2[j] start from 'j'
* c) 'k' is the length of substring
*
* Initialization
*
* dp[1][i][j] = (s1[i] == s2[j] ? true : false)
*
* Formula
*
* same as the above recursive method idea
*
* dp[k][i][j] =
* dp[divk][i][j] && dp[k-divk][i+divk][j+divk] ||
* dp[divk][i][j+k-divk] && dp[k-divk][i+divk][j]
*
* `divk` mean split the k to two parts, so 0 <= divk <= k;
*/
bool
isScramble_dp
(string s1, string s2) {
if
(s1.
size
()!= s2.
size
() || s1.
size
()==
0
|| s2.
size
()==
0
) {
return
false
;
}
if
(s1 == s2){
return
true
;
}
const
int
len = s1.
size
();
//
dp[len+1][len][len]
vector< vector< vector<
bool
> > >
dp
(len+
1
, vector< vector<
bool
> >(len, vector<
bool
>(len) ) );
//
ignor the k=0, just for readable code.
//
initialization k=1
for
(
int
i=
0
; i<len; i++){
for
(
int
j=
0
; j<len; j++) {
dp[
1
][i][j] = (s1[i] == s2[j]);
}
}
//
start from k=2 to len, O(n^4) loop.
for
(
int
k=
2
; k<=len; k++){
for
(
int
i=
0
; i<len-k+
1
; i++){
for
(
int
j=
0
; j<len-k+
1
; j++){
dp[k][i][j] =
false
;
for
(
int
divk =
1
; divk < k && dp[k][i][j]==
false
; divk++){
dp[k][i][j] = ( dp[divk][i][j] && dp[k-divk][i+divk][j+divk] ) ||
( dp[divk][i][j+k-divk] && dp[k-divk][i+divk][j] );
}
}
}
}
return
dp[len][
0
][
0
];
}
bool
isScramble
(string s1, string s2) {
srand
(
time
(
0
));
if
(
random
()%
2
) {
cout <<
"
---- recursion ---
"
<< endl;
return
isScramble_recursion
(s1, s2);
}
cout <<
"
---- dynamic programming ---
"
<< endl;
return
isScramble_dp
(s1, s2);
}
int
main
(
int
argc,
char
** argv)
{
string s1=
"
great
"
, s2=
"
rgtae
"
;
if
(argc>
2
){
s1 = argv[
1
];
s2 = argv[
2
];
}
cout << s1 <<
"
,
"
<< s2 << endl;
cout <<
isScramble
(s1, s2) << endl;
return
0
;
}
Back
|
FazBrowse Home
|
New Git URL