97. Interleaving String

/*dp[i][j] 代表s3前i+j个字符 是否能够用s1的前 i个字符 和s2的前j个字符 表示i j 代表字符的个数 不是下标dp[0][0] =truedp[i][0]=(dp[i-1][0]&&s1[i-1]==s3[i-1])dp[0][j]=(dp[0][j-1]&&s2[j-1]==s3[j-1])dp[i][j]=(dp[i-1][j]&&s1[i-1]==s3[i+j-1])||(dp[i][j-1]&&s2[j-1]==s3[i+j-1])*/class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        if(s1.length()+s2.length()!=s3.length())            return false;        vector<vector<bool>> dp(s1.length()+1,vector<bool>(s2.length()+1,false));        for(int i=0;i<=s1.length();i++)            for(int j=0;j<=s2.length();j++)            {                if(i==0&&j==0)                    dp[i][j]=true;                else if(i==0)                    dp[0][j]=dp[0][j-1]&&s2[j-1]==s3[j-1];                else if(j==0)                    dp[i][0]=dp[i-1][0]&&s1[i-1]==s3[i-1];                else                     dp[i][j]=(dp[i-1][j]&&s1[i-1]==s3[i+j-1])||(dp[i][j-1]&&s2[j-1]==s3[i+j-1]);            }        return dp[s1.length()][s2.length()];    }};