[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.

First we can consider this as a recursive problem.
If  s3[i1+i2+1] == s1[i1+1], search(i1+1,i2)
If  s3[i1+i2+1] == s2[i2+1], search(i1,i2+1)
until end. However, this can only pass the small test cases, but failed on the large test cases.

So we need to think of the dynamic programming (DP), which usually have much less complexity. In this problem, a 2D DP is more suitable.  As usual, the typical way of solving dp is to find the state, and the optimal function. Here, the state can be considered as: A[i][j], which means S3[i+j] can be formed by S1[i] and S2[j] (for simplicity here string starts from 1, in the code we need to deal with that string starts from 0).

So, we have the optimal function:
  A[i][j] =   (s3[i+j]==s1[i]  && match[i-1][j])  || (s3[i+j] ==s2[j] && match[i][j-1])

c++

bool isInterleave(string s1, string s2, string s3) {         if(s1.size()+s2.size() != s3.size()) return false;    bool **dp = new bool *[s1.size()+1];    for(int i=0; i<s1.size()+1; i++){        dp[i] = new bool [s2.size()+1];    }    dp[0][0] = true;    for(int i=1; i<s1.size()+1; i++){        if(s1[i-1]==s3[i-1] && dp[i-1][0])            dp[i][0] = true;    }    for(int j=1; j<s2.size()+1; j++){        if(s2[j-1]==s3[j-1] && dp[0][j-1])            dp[0][j] = true;    }    for(int i=1; i<s1.size()+1; i++){        for(int j=1; j<s2.size()+1; j++){            if(s1[i-1]==s3[i+j-1] && dp[i-1][j])                dp[i][j] = true;            if(s2[j-1]==s3[i+j-1] && dp[i][j-1])                dp[i][j] = true;        }    }    return dp[s1.size()][s2.size()];    }

java

public boolean isInterleave(String s1, String s2, String s3) {        int n1 = s1.length();        int n2 = s2.length();        if(n1+n2!=s3.length()) return false;        boolean [][]flag = new boolean [n1+1][n2+1];        for(int i=0;i<=n1;i++){        for(int j=0;j<=n2;j++){        flag[i][j] = false;        }        }        flag[0][0] = true;        for(int i=1;i<=n1;i++){        if(s1.charAt(i-1)==s3.charAt(i-1)&& flag[i-1][0]) flag[i][0] = true;        }        for(int j=1;j<=n2;j++){        if(s2.charAt(j-1)==s3.charAt(j-1)&& flag[0][j-1]) flag[0][j] =true;        }        for(int i=1;i<=n1;i++){        for(int j=1;j<=n2;j++){        if((flag[i-1][j] && s1.charAt(i-1)==s3.charAt(i-1+j))||        (flag[i][j-1]&& s2.charAt(j-1)==s3.charAt(i+j-1))){        flag[i][j] = true;        }        }        }        return flag[n1][n2];    }