[leetcode] Interleaving String@DP

Given s1, s2, s3,
find whether s3 is formed by the interleaving of s1 and s2.

For example,

Given:
s1 = 

"aabcc"

,
s2 = 

"dbbca"

,

When s3 = 

"aadbbcbcac"

,
return true.

When s3 = 

"aadbbbaccc"

,
return false.

首先说一下这个interleaving的意思，这个单词的意思就是s3是否能通过s1和s2交织得到，s1，s2中字符的相对顺序是不能被打乱的。

递归的解法如下：

class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return _isInterleave(s1,s2,s3);
}
bool _isInterleave(string s1,string s2,string s3){
int length1=s1.size();
int length2=s2.size();
int length3=s3.size();
if(length1+length2!=length3)
return false;
if(s3=="")
return s1==""&&s2=="";
if(s1=="" && s2=="")
return s3=="";
if(s1=="")
return s2==s3;
if(s2=="")
return s1==s3;
string s11(s1.begin()+1,s1.end());
string s21(s2.begin()+1,s2.end());
string s31(s3.begin()+1,s3.end());
if(s1[0]==s3[0] && s2[0]!=s3[0])
return _isInterleave(s11,s2,s31);
else if(s1[0]!=s3[0] && s2[0]==s3[0])
return _isInterleave(s1,s21,s31);
else if(s1[0]==s3[0] && s2[0]==s2[0])
return _isInterleave(s11,s2,s31)||_isInterleave(s1,s21,s31);
else
return false;
}
};

非递归的解法如下：

class Solution {
private:
public:
bool isInterleave(string s1, string s2, string s3) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int length1=s1.size();
int length2=s2.size();
int length3=s3.size();
if (length1 + length2 != length3)
return false;
bool (*f)[1000]=new bool[s1.size()+1][1000];
f[0][0] = true;
for(int i = 1; i <= length1; i++)
f[i][0] = f[i-1][0] && (s3[i-1] == s1[i-1]);

for(int j = 1; j <= length2; j++)
f[0][j] = f[0][j-1] && (s3[j-1] == s2[j-1]);

for(int i = 1; i <= length1; i++)
for(int j = 1; j <= length2; j++)
f[i][j] = (f[i][j-1] && s2[j-1] == s3[i+j-1]) || (f[i-1][j] && s1[i-1] == s3[i+j-1]);

return f[length1][length2];
}
};