97. 交错字符串(动态规划)

给定三个字符串 s1, s2, s3, 验证 s3 是否是由 s1 和 s2 交错组成的。

示例 1:

输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出: true

示例 2:

输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出: false

示例 :

输入: s1 = "", s2 = "", s3 = ""
输出: true

示例 4:

输入: s1 = "abc", s2 = "", s3 = "abc"
输出: true

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {

        if(s1.size()+s2.size()!=s3.size())
        return false;
     vector<vector<bool>>dp(s1.size()+1,vector<bool>(s2.size()+1,false));
     dp[0][0]=true;
     for(int i=0;i<=s1.size();i++){
         for(int j=0;j<=s2.size();j++){
             if(i==0&&j==0)
                dp[i][j]=true;
            else if(i==0)
                dp[i][j]=dp[i][j-1]&&s2[j-1]==s3[i+j-1];
            else if(j==0)
                dp[i][j]=(dp[i-1][j]&&s1[i-1]==s3[i+j-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]);
            //  cout<<dp[i][j]<<endl;
         }
     }
      return dp[s1.size()][s2.size()];
    }
   
};