LeetCode 97 Interleaving String(Python详解及实现)

【题目】

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.

 

给定三个字符串S1，S2，S3，判断S3是否能够由S1，S2交错组合而成。

 

【思路】

二维动态规划：

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



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。dp[i][j]就表示s1的前i个和s2的前j个是否和s3的前i+j个匹配成功

标1的匹配成功。

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]。

 

【Python实现】

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]