leetcode_[python/C++]_97_Interleaving String_动态规划

题目链接

【题目】

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.

Recurse:

Judge Small: Accepted!

Judge Large: Time Limit Exceeded

(1)递归的思想，(Time error)

class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
if(s1.length() == 0)
return s2 == s3;
if(s2.length() == 0)
return s1 == s3;
if(s3.length() == 0)
return s1.length() + s2.length() == 0;
if(s1[0] == s3[0] && s2[0] != s3[0])
return isInterleave(s1.substr(1),s2,s3.substr(1));
else if(s1[0] != s3[0] && s2[0] == s3[0])
return isInterleave(s1,s2.substr(1),s3.substr(1));
else if(s1[0] == s3[0] && s2[0] == s3[0])
return isInterleave(s1,s2.substr(1),s3.substr(1)) || isInterleave(s1.substr(1),s2,s3.substr(1));
else
return false;
}
};

(2)动态规划：

建立二维数组，可达为1,不可达为0，不断往右往下，如果能走到右下角，则return true

s1 = “aabcc”

s2 = “dbbca”

s3 = “aadbbcbcac”

class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
int size1 = s1.length();
int size2 = s2.length();
int size3 = s3.length();
if(size1 + size2 != size3)
return false;
int dp[size1+1][size2+1] = {0};
for( int i = 0 ; i <= size1 ; i ++ )
for( int j = 0 ; j <= size2 ; j ++ )
{
if( i == 0 && j == 0 )
dp[i][j] = 1;
else if( i > 0 && dp[i-1][j] && s3[i+j-1] == s1[i-1])
dp[i][j] = 1;
else if( j > 0 && dp[i][j-1] && s3[i+j-1] == s2[j-1])
dp[i][j] = 1;
else
dp[i][j] = 0;
}
return dp[size1][size2];
}
};

python

动态规划：

class Solution(object):
def isInterleave(self, s1, s2, s3):
size1 = len(s1)
size2 = len(s2)
size3 = len(s3)
if size1 + size2 != size3:
return False
dp = [[0]*(size2 + 1) for i in range(size1 + 1)]
for i in range(size1 + 1):
for j in range(size2 + 1):
if i == 0 and j == 0:
dp[i][j] = 1
elif i > 0 and dp[i-1][j] and s3[i + j - 1] == s1[i - 1]:
dp[i][j] = 1
elif j > 0 and dp[i][j -1] and s3[i + j - 1] == s2[j - 1]:
dp[i][j] = 1
else:
dp[i][j] = 0
return True if dp[size1][size2] else False