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ScrambleString.cpp
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ScrambleString.cpp
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/*
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.
*/
class Solution {
public:
bool isScramble(string s1, string s2) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int slen=s1.size();
if(slen!=s2.size())
return false;
vector<vector<vector<unsigned char> > > matrix;
for(int i=0;i<slen;i++)
matrix.push_back(vector<vector<unsigned char> >(slen-i,vector<unsigned char>(slen-i,0x0)));
for(int i=0;i<slen;i++){
for(int j=0;j<slen;j++){
if(s1[i]==s2[j])
matrix[0][i][j]=1;
}
}
for(int i=1;i<slen;i++){
for(int j=0;j<slen-i;j++){
for(int k=0;k<slen-i;k++){
for(int l=0;l<i;l++){
if(
(matrix[l][j][k]&&matrix[i-l-1][j+l+1][k+l+1])||
(matrix[l][j][k+i-l]&&matrix[i-l-1][j+l+1][k])
){
matrix[i][j][k]=1;
break;
}
}
}
}
}
return matrix[slen-1][0][0];
}
};