# [LeetCode] 97. Interleaving String 交织相错的字符串

2021年09月15日 阅读数：1

Given s1s2s3, find whether s3 is formed by the interleaving of s1and s2.html

Example 1:java

```Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
```

Example 2:python

```Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false```

dp[i][j] = (dp[i - 1][j] && s1[i - 1] == s3[i - 1 + j]) || (dp[i][j - 1] && s2[j - 1] == s3[j - 1 + i]);spa

Java:htm

```public boolean isInterleave(String s1, String s2, String s3) {

if ((s1.length()+s2.length())!=s3.length()) return false;

boolean[][] matrix = new boolean[s2.length()+1][s1.length()+1];

matrix[0][0] = true;

for (int i = 1; i < matrix[0].length; i++){
matrix[0][i] = matrix[0][i-1]&&(s1.charAt(i-1)==s3.charAt(i-1));
}

for (int i = 1; i < matrix.length; i++){
matrix[i][0] = matrix[i-1][0]&&(s2.charAt(i-1)==s3.charAt(i-1));
}

for (int i = 1; i < matrix.length; i++){
for (int j = 1; j < matrix[0].length; j++){
matrix[i][j] = (matrix[i-1][j]&&(s2.charAt(i-1)==s3.charAt(i+j-1)))
|| (matrix[i][j-1]&&(s1.charAt(j-1)==s3.charAt(i+j-1)));
}
}

return matrix[s2.length()][s1.length()];

}
```

Python:blog

```# O(m*n) space
def isInterleave1(self, s1, s2, s3):
r, c, l= len(s1), len(s2), len(s3)
if r+c != l:
return False
dp = [[True for _ in xrange(c+1)] for _ in xrange(r+1)]
for i in xrange(1, r+1):
dp[i][0] = dp[i-1][0] and s1[i-1] == s3[i-1]
for j in xrange(1, c+1):
dp[0][j] = dp[0][j-1] and s2[j-1] == s3[j-1]
for i in xrange(1, r+1):
for j in xrange(1, c+1):
dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i-1+j]) or \
(dp[i][j-1] and s2[j-1] == s3[i-1+j])
return dp[-1][-1]
```

Python:

```# O(2*n) space
def isInterleave2(self, s1, s2, s3):
l1, l2, l3 = len(s1)+1, len(s2)+1, len(s3)+1
if l1+l2 != l3+1:
return False
pre = [True for _ in xrange(l2)]
for j in xrange(1, l2):
pre[j] = pre[j-1] and s2[j-1] == s3[j-1]
for i in xrange(1, l1):
cur = [pre[0] and s1[i-1] == s3[i-1]] * l2
for j in xrange(1, l2):
cur[j] = (cur[j-1] and s2[j-1] == s3[i+j-1]) or \
(pre[j] and s1[i-1] == s3[i+j-1])
pre = cur[:]
return pre[-1]
```

Python:

```# O(n) space
def isInterleave3(self, s1, s2, s3):
r, c, l= len(s1), len(s2), len(s3)
if r+c != l:
return False
dp = [True for _ in xrange(c+1)]
for j in xrange(1, c+1):
dp[j] = dp[j-1] and s2[j-1] == s3[j-1]
for i in xrange(1, r+1):
dp[0] = (dp[0] and s1[i-1] == s3[i-1])
for j in xrange(1, c+1):
dp[j] = (dp[j] and s1[i-1] == s3[i-1+j]) or (dp[j-1] and s2[j-1] == s3[i-1+j])
return dp[-1]
```

Python:

```# DFS
def isInterleave4(self, s1, s2, s3):
r, c, l= len(s1), len(s2), len(s3)
if r+c != l:
return False
stack, visited = [(0, 0)], set((0, 0))
while stack:
x, y = stack.pop()
if x+y == l:
return True
if x+1 <= r and s1[x] == s3[x+y] and (x+1, y) not in visited:
if y+1 <= c and s2[y] == s3[x+y] and (x, y+1) not in visited:
return False
```

Python:

```# BFS
def isInterleave(self, s1, s2, s3):
r, c, l= len(s1), len(s2), len(s3)
if r+c != l:
return False
queue, visited = [(0, 0)], set((0, 0))
while queue:
x, y = queue.pop(0)
if x+y == l:
return True
if x+1 <= r and s1[x] == s3[x+y] and (x+1, y) not in visited:
if y+1 <= c and s2[y] == s3[x+y] and (x, y+1) not in visited:
return False
```

Python:

```# Time:  O(m * n)
# Space: O(m + n)
class Solution(object):
# @return a boolean
def isInterleave(self, s1, s2, s3):
if len(s1) + len(s2) != len(s3):
return False
if len(s1) > len(s2):
return self.isInterleave(s2, s1, s3)
match = [False for i in xrange(len(s1) + 1)]
match[0] = True
for i in xrange(1, len(s1) + 1):
match[i] = match[i -1] and s1[i - 1] == s3[i - 1]
for j in xrange(1, len(s2) + 1):
match[0] = match[0] and s2[j - 1] == s3[j - 1]
for i in xrange(1, len(s1) + 1):
match[i] = (match[i - 1] and s1[i - 1] == s3[i + j - 1]) \
or (match[i] and s2[j - 1] == s3[i + j - 1])
return match[-1]
```

Python:

```# Time:  O(m * n)
# Space: O(m * n)
# Dynamic Programming
class Solution2(object):
# @return a boolean
def isInterleave(self, s1, s2, s3):
if len(s1) + len(s2) != len(s3):
return False
match = [[False for i in xrange(len(s2) + 1)] for j in xrange(len(s1) + 1)]
match[0][0] = True
for i in xrange(1, len(s1) + 1):
match[i][0] = match[i - 1][0] and s1[i - 1] == s3[i - 1]
for j in xrange(1, len(s2) + 1):
match[0][j] = match[0][j - 1] and s2[j - 1] == s3[j - 1]
for i in xrange(1, len(s1) + 1):
for j in xrange(1, len(s2) + 1):
match[i][j] = (match[i - 1][j] and s1[i - 1] == s3[i + j - 1]) \
or (match[i][j - 1] and s2[j - 1] == s3[i + j - 1])
return match[-1][-1]
```

Python:

```# Time:  O(m * n)
# Space: O(m * n)
# Recursive + Hash
class Solution3(object):
# @return a boolean
def isInterleave(self, s1, s2, s3):
self.match = {}
if len(s1) + len(s2) != len(s3):
return False
return self.isInterleaveRecu(s1, s2, s3, 0, 0, 0)

def isInterleaveRecu(self, s1, s2, s3, a, b, c):
if repr([a, b]) in self.match.keys():
return self.match[repr([a, b])]

if c == len(s3):
return True

result = False
if a < len(s1) and s1[a] == s3[c]:
result = result or self.isInterleaveRecu(s1, s2, s3, a + 1, b, c + 1)
if b < len(s2) and s2[b] == s3[c]:
result = result or self.isInterleaveRecu(s1, s2, s3, a, b + 1, c + 1)

self.match[repr([a, b])] = result

return result```

C++:

```public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if (s1.length() + s2.length() != s3.length()) return false;
int n1 = s1.length(), n2 = s2.length();
boolean[][] dp = new boolean[n1 + 1][n2 + 1];
for (int i = 0; i <= n1; i++) {
for (int j = 0; j <= n2; j++) {
if (i == 0 && j == 0) { // s1 empty, s2 empty
dp[i][j] = true;
} else {
dp[i][j] = (i > 0 && dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1)) || (j > 0 && dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1));
}
}
}
return dp[n1][n2];
}
}
```

[LeetCode] 139. Word Break 单词拆分

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