[LeetCode] Interleaving String 交叉字符串

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.

    bool isInterleave(string s1, string s2, string s3) {                // if length does not match        if((s1.length()+s2.length())!=s3.length())            return false;                if(s1.length()==0)            if(s2.compare(s3)==0)                return true;            else                return false;                        if(s2.length()==0)            if(s1.compare(s3)==0)                return true;             else                return false;                // if the first char does not match        if(s1.at(0)!=s3.at(0) && s2.at(0)!=s3.at(0))            return false;                // if the first char of s1 matches with the first char of s3        if(s1.at(0)==s3.at(0) && s2.at(0)!=s3.at(0) )            return isInterleave(s1.substr(1,s1.length()-1), s2, s3.substr(1, s3.length()-1) );                // if the first char of s2 matches with the first char of s3        if( s1.at(0)!=s3.at(0) && s2.at(0)==s3.at(0) )            return isInterleave(s1, s2.substr(1,s2.length()-1), s3.substr(1, s3.length()-1) );                    // if the first char of s1 and s2 matches with the first char of s3        if(s1.at(0)==s3.at(0) && s2.at(0)==s3.at(0))            return (isInterleave(s1.substr(1,s1.length()-1), s2, s3.substr(1, s3.length()-1) ) ) ||( isInterleave(s1, s2.substr(1,s2.length()-1), s3.substr(1, s3.length()-1) ) );    }

在Leetcode的服务器上，在测试非常长的字符串时，上面的代码给出了超时的错误。

可以考虑用动态规划的思想，把上面递归的中间结果存在一个表中。代码如下：

    bool isInterleave(string s1, string s2, string s3) {                if((s1.length()+s2.length())!=s3.length())            return false;                vector<bool> table_row (s2.length()+1, false);        vector<vector<bool>> table(s1.length()+1, table_row ); // this table stores the intermediate results        for(int i=0; i<=s1.length(); i++)            if(s1.substr(0,i).compare(s3.substr(0,i)) ==0)                table[i][0] = true; // initialize the table                for(int j=0; j<=s2.length(); j++)            if(s2.substr(0,j).compare(s3.substr(0,j)) ==0)                table[0][j] = true; // initialize the table                for(int i=1; i<=s1.length(); i++)            for(int j=1; j<=s2.length(); j++)            {                if( s1.at(i-1)==s3.at(i+j-1) && s2.at(j-1)!=s3.at(i+j-1) )                    table[i][j] = table[i-1][j];                                    if( s1.at(i-1)!=s3.at(i+j-1) && s2.at(j-1)==s3.at(i+j-1) )                    table[i][j] = table[i][j-1];                                    if( s1.at(i-1)==s3.at(i+j-1) && s2.at(j-1)==s3.at(i+j-1) )                    table[i][j] = table[i-1][j] || table[i][j-1];            }                    return table[s1.length()][s2.length()];    }