Leetcode--Interleaving String

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 function(string s1,int flag1,string s2,int flag2,string s3,int flag3)    {        if(flag3==s3.size())        {            if(flag2==s2.size()&&flag1==s1.size())                return true;            else                 return false;        }        else if(flag3<s3.size())        {            if((flag1<s1.size()&&s1[flag1]==s3[flag3])&&(flag2<s2.size()&&s2[flag2]==s3[flag3]))                return function(s1,flag1+1,s2,flag2,s3,flag3+1)||function(s1,flag1,s2,flag2+1,s3,flag3+1);            else if(flag1<s1.size()&&s1[flag1]==s3[flag3])                return function(s1,flag1+1,s2,flag2,s3,flag3+1);            else if(flag2<s2.size()&&s2[flag2]==s3[flag3])                return function(s1,flag1,s2,flag2+1,s3,flag3+1);            else                return false;        }    }    bool isInterleave(string s1, string s2, string s3)    {        return function(s1,0,s2,0,s3,0);    }

结果超时

Solution2: 利用动态规划

定义arr[i][j]表示能否用s1的前i个字符和s2的前j个字符所产生前i+j位s3

arr[0][0]=true

最后的结果是arr[s1.size()][s2.size()]

时间复杂度为O(s1.size()*s2.size())

 bool isInterleave(string s1, string s2, string s3)    {        if(s1.size()+s2.size()!=s3.size())//s3的长度不等于s1和s之和            return false;                    bool **arr=new bool*[s1.size()+1];//申请二维空间        for(int i=0;i<s1.size()+1;i++)            arr[i]=new bool[s2.size()+1];                arr[0][0]=true;//初始化            for(int i=1;i<=s1.size();i++)        {            arr[i][0]=arr[i-1][0]&&s1[i-1]==s3[i-1];        }                for(int j=1;j<=s2.size();j++)            arr[0][j]=arr[0][j-1]&&s2[j-1]==s3[j-1];                                for(int i=1;i<=s1.size();i++)            for(int j=1;j<=s2.size();j++)                arr[i][j]=false;                        for(int i=1;i<=s1.size();i++)//用递推式计算arr[i][j]的值            for(int j=1;j<=s2.size();j++)            {                    arr[i][j]=arr[i-1][j]&&s1[i-1]==s3[i+j-1];//s3的最后一个字符是来自s1                                        arr[i][j]=arr[i][j]||(arr[i][j-1]&&s2[j-1]==s3[i+j-1]);//s3的最后一个字符是来自s2            }                    return arr[s1.size()][s2.size()];    }