------
## 最终结论
```python
def isInterleave(s1, s2, s3):
    if len(s1) + len(s2) != len(s3):
        return False
    
    dp = [[False] * (len(s2) + 1) for _ in range(len(s1) + 1)]
    dp[0][0] = True
    
    for i in range(1, len(s1) + 1):
        dp[i][0] = dp[i-1][0] and s1[i-1] == s3[i-1]
    
    for j in range(1, len(s2) + 1):
        dp[0][j] = dp[0][j-1] and s2[j-1] == s3[j-1]
    
    for i in range(1, len(s1) + 1):
        for j in range(1, len(s2) + 1):
            dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i+j-1]) or (dp[i][j-1] and s2[j-1] == s3[i+j-1])
    
    return dp[len(s1)][len(s2)]

# Example usage:
s1 = "aabcc"
s2 = "dbbca"
s3 = "aadbbcbcac"
print(isInterleave(s1, s2, s3))  # Output: True

s3 = "aadbbbaccc"
print(isInterleave(s1, s2, s3))  # Output: False
```