【題目】

Given s1, s2, s3, find whether s3 is formedby the interleaving of s1 and s2.

For example,

Given:

s1 = "aabcc",

s2 = "dbbca",

When s3 = "aadbbcbcac", returntrue.

When s3 = "aadbbbaccc", returnfalse.

給定三個字串S1，S2，S3，判斷S3是否能夠由S1，S2交錯組合而成。

【思路】

二維動態規劃：

s1 = "aabcc"  s2 = "dbbca"  s3 = "aadbbcbcac"

s1, s2只有兩個字串，因此可以展平為一個二維地圖，判斷是否能從左上角走到右下角。dp[i][j]就表示s1的前i個和s2的前j個是否和s3的前i+j個匹配成功

標1的匹配成功。

當s1到達第i個元素，s2到達第j個元素:

地圖上往右一步就是s2[ j-1]匹配s3[i+j-1]。

地圖上往下一步就是s1[ i-1]匹配s3[i+j-1]。

【Python實現】

class Solution(object):
    def isInterleave(self, s1, s2, s3):
        """
        :type s1: str
        :type s2: str
        :type s3: str
        :rtype: bool
        """
        len_s1 = len(s1)
        len_s2 = len(s2)
        len_s3 = len(s3)
        if len_s1 + len_s2 != len_s3:
            return False
        dp = [[False for i in range(len_s2 + 1)] for j in range(len_s1 + 1)]
        for i in range(0,len_s1 + 1):
            for j in range(0,len_s2 + 1):
                if i == 0 and j == 0:
                    dp[i][j] = True
                elif i == 0:
                    dp[i][j] = dp[i][j-1] and (s2[j-1] == s3[j-1])
                elif j == 0:
                    dp[i][j] = dp[i-1][j] and (s1[i-1] == s3[i-1])
                else:
                    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]