描述
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.
分析
二维动态规划的思路，令dp[i][j]表示s3的前i + j个字符是否可以由s1的前i个和s2的前j个字符交织组成，显然有dp[0][0] = True，而dp[0][j]和dp[i][0]可以很容易预先填充，然后对于dp[i][j]，有递推公式
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])
代码
Python
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class (object):
def isInterleave(self, s1, s2, s3):
"""
:type s1: str
:type s2: str
:type s3: str
:rtype: bool
"""
n1 = len(s1)
n2 = len(s2)
if len(s3) != n1 + n2:
return False
dp = [[False] * (n2 + 1) for i in xrange(n1 + 1)]
dp[0][0] = True
for i in xrange(1, n1 + 1):
dp[i][0] = dp[i - 1][0] and s1[i - 1] == s3[i - 1]
if not dp[i][0]:
break
for j in xrange(1, n2 + 1):
dp[0][j] = dp[0][j - 1] and s2[j - 1] == s3[j - 1]
if not dp[0][j]:
break
for i in xrange(1, n1 + 1):
for j in xrange(1, n2 + 1):
if dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]:
dp[i][j] = True
if dp[i][j - 1] and s2[j - 1] == s3[i + j - 1]:
dp[i][j] = True
return dp[n1][n2]
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