交叉字符串

class Solution:
    """
    @param s1: A string
    @param s2: A string
    @param s3: A string
    @return: Determine whether s3 is formed by interleaving of s1 and s2
    """
    def isInterleave(self, s1, s2, s3):
        # write your code here
        n = len(s1)
        m = len(s2)
        if m+n != len(s3):
            return False
        if n == 0:
            return s3 == s2
        if m== 0:
            return s3 == s1
        if n == 1:
            for i in range(m+1):
                if s3[i] == s1:
                    return s3[:i]+s3[i+1:] ==s2
                   
        if m == 1:
            for i in range(n+1):
                if s3[i] == s2:
                    return s3[:i] + s3[i+1:] == s1

            
        lt = [[0 for i in range(n+1)] for j in range(m+1)] #m hang n lie 
        lt[0][0] =True
        
        for i in range(1,m+1):
            lt[i][0] = (lt[i-1][0] and s2[i-1] == s3[i-1])
            
        for j in range(1,n+1):
            lt[0][j] = (lt[0][j-1] and s1[j-1] == s3[j-1])
            
        for i in range(1,m+1):
            for j in range(1,n+1):
                lt[i][j] = (lt[i-1][j] and s2[i-1] == s3[i+j-1]) or (lt[i][j-1] and s1[j-1] == s3[i+j-1])

        
        return lt[m][n]
        
        

给出三个字符串:s1、s2、s3，判断s3是否由s1和s2交叉构成。

 

样例

比如 s1 = "aabcc" s2 = "dbbca"

    - 当 s3 = "aadbbcbcac"，返回  true.

    - 当 s3 = "aadbbbaccc"， 返回 false.

挑战

要求时间复杂度为O(n^2)或者更好

 

采用动态规划的思想，参考的是邹博算法教材！（注意第一行和第一列的初始化）