97. Interleaving String【H】【73】

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.

Subscribe to see which companies asked this question

可以用递归做，每匹配s1或者s2中任意一个就递归下去。但是会超时。

因此考虑用动态规划做。

s1, s2只有两个字符串，因此可以展平为一个二维地图，判断是否能从左上角走到右下角。

当s1到达第i个元素，s2到达第j个元素:

地图上往右一步就是s2[j-1]匹配s3[i+j-1]。

地图上往下一步就是s1[i-1]匹配s3[i+j-1]。

示例：s1="aa",s2="ab",s3="aaba"。标1的为可行。最终返回右下角。

     0  a  b

0   1  1  0

a   1  1  1

a   1  0  1 

class Solution(object):
    def isInterleave(self, s1, s2, s3):
        if s3 == '':
            return True
        if len(s1) + len(s2) != len(s3):
            return False
        if s1 == '' or s2 == '':
            return s3 == s1 + s2

        dp = []
        for i in xrange(len(s2)+1):
            dp += [0] * (len(s1)+1),

        if s1[0] == s3[0] or s2[0] == s3[0]:
            dp[0][0] = 1

        for i in xrange(len(s1)):
            if (s1[i] == s3[i]) and dp[0][i] == 1:
                dp[0][i+1] = 1
        for i in xrange(len(s2)):
            if s2[i] == s3[i] and dp[i][0] == 1:
                dp[i+1][0] = 1

        for i in xrange(len(s2)):
            for j in xrange(len(s1)):
                #print i,j,s1[j],s2[i],s3[i+j+1]
                if (dp[i+1][j] == 1 and s1[j] == s3[i+j+1]) or (dp[i][j+1] == 1 and s2[i] == s3[i+j+1]):
                    dp[i+1][j+1] = 1

        return dp[-1][-1] == 1

        """
        :type s1: str
        :type s2: str
        :type s3: str
        :rtype: bool
        """