leetcode || 97、Interleaving String

problem：

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.

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Dynamic Programming String

题意：能否 用两个字符串顺序穿插成第三个字符串

thinking：

（1）首先想到用递归，提交超时。

（2）这道题正确解法是 二维DP，状态转移方程：

很牛的解题：http://www.cnblogs.com/remlostime/archive/2012/11/25/2787297.html

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]);

只有 s3的第 i+i-1 个字符与s1的第i-1个字符 或 s2的第j-1个字符相同时，才有可能有效。还要参考前面的结果。

code：

DP：

class Solution {
private:
bool f[1000][1000];
public:
bool isInterleave(string s1, string s2, string s3) {
if (s1.size() + s2.size() != s3.size())
return false;

f[0][0] = true;
for(int i = 1; i <= s1.size(); i++)  //f[i][0]初始化
f[i][0] = f[i-1][0] && (s3[i-1] == s1[i-1]);

for(int j = 1; j <= s2.size(); j++)  //f[0][j]初始化
f[0][j] = f[0][j-1] && (s3[j-1] == s2[j-1]);

for(int i = 1; i <= s1.size(); i++)
for(int j = 1; j <= s2.size(); 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[s1.size()][s2.size()];
}
};

递归：超时

class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
s1.push_back('0');
s2.push_back('0');
s3.push_back('0');
bool flag=false;
check(s1,0,s2,0,s3,0,flag);
return flag;

}
protected:
void check(string s1,int i,string s2,int j,string s3,int k, bool &flag)
{
if(flag)
return;
if(s1[i]=='0'&&s2[j]=='0'&&s3[k]=='0')
{
flag=true;
return;
}
if(s1[i]=='0'&&s2[j]=='0'&&s3[k]!='0')
return;
if((s1[i]!='0'&&s2[j]=='0'&&s3[k]=='0')||(s1[i]=='0'&&s2[j]!='0'&&s3[k]=='0'))
return;
if(s1[i]!=s3[k]&&s2[j]!=s3[k])
return;
if(s1[i]==s3[k])
check(s1,i+1,s2,j,s3,k+1,flag);
if(s2[j]==s3[k])
check(s1,i,s2,j+1,s3,k+1,flag);
}
};