原题地址:https://oj.leetcode.com/problems/interleaving-string/
题意:
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.
解题思路:动态规划。dp[i][j]表示s1[0...i-1]和s2[0...j-1]是否可以拼接为s3[0...i+j-1],可以拼接为true,不可以拼接为false。
代码:
class Solution:
# @return a boolean
def isInterleave(self, s1, s2, 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)]
dp[0][0]=True
for i in range(1,len(s1)+1):
dp[i][0] = dp[i-1][0] and s3[i-1]==s1[i-1]
for i in range(1,len(s2)+1):
dp[0][i] = dp[0][i-1] and s3[i-1]==s2[i-1]
for i in range(1,len(s1)+1):
for j in range(1,len(s2)+1):
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)]
[leetcode]Interleaving String @ Python,布布扣,bubuko.com
时间: 2024-10-13 06:33:09