97. 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.

• 这道题目比较有难度。当然用递归的方法比较简单，但是会出现超时的问题，非递归用dp比较有难度，不太容易想明白递归公式。

递归方法：bool isInterleave(char* s1, char* s2, char* s3) {    int n1 = strlen(s1);    int n2 = strlen(s2);    int n3 = strlen(s3);    if((n1+n2)!=n3){        return false;    }    if(n3 == 0 && n1 == 0 && n2 == 0){        return true;    }    if(s1[0] != s3[0] && s2[0] != s3[0]){        return false;    }    if(s1[0] != '\0' && s1[0] == s3[0] && s2[0]!=s3[0]){        return isInterleave(s1+1,s2,s3+1);    }    if(s2[0] != '\0' && s1[0] != s3[0] && s2[0]==s3[0]){        return isInterleave(s1,s2+1,s3+1);    }    if(s1[0] == s3[0] && s2[0] == s3[0]){        return isInterleave(s1,s2+1,s3+1)||               isInterleave(s1+1,s2,s3+1);    }    return true;}

非递归方法class Solution {public:            bool isInterleave(string s1, string s2, string s3) {        int n1 = s1.size();        int n2 = s2.size();        int n3 = s3.size();        vector<bool> t(n2+1,false);        vector<vector<bool>> dp(n1+1,t);        if(n1+n2 != n3){            return false;        }        for(int i = 0; i <= n1; ++i){            for(int j = 0; j<= n2; ++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][j-1] && s2[j-1] == s3[i+j-1])||(dp[i-1][j] && s1[i-1] == s3[i+j-1]);                }                        }        }        return dp[n1][n2];    }};