dp/interleaving-string
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2023-12-01
Interleaving String
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
设状态f[i][j]
,表示s1[0,i]
和s2[0,j]
,匹配s3[0, i+j]
。如果 s1 的最后一个字符等于 s3 的最后一个字符,则f[i][j]=f[i-1][j]
;如果 s2 的最后一个字符等于 s3 的最后一个字符,则f[i][j]=f[i][j-1]
。因此状态转移方程如下:
f[i][j] = (s1[i - 1] == s3 [i + j - 1] && f[i - 1][j])
|| (s2[j - 1] == s3 [i + j - 1] && f[i][j - 1]);
递归
// Interleaving String
// 递归,会超时,仅用来帮助理解
public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if (s3.length() != s1.length() + s2.length())
return false;
if (s1.isEmpty() || s2.isEmpty()) return true;
return isInterleave(s1, 0, s1.length(),
s2, 0, s2.length(), s3, 0, s3.length());
}
private static boolean isInterleave(String s1, int begin1, int end1,
String s2, int begin2, int end2,
String s3, int begin3, int end3) {
if (begin3 == end3)
return begin1 == end1 && begin2 == end2;
return (begin1 < end1 && s1.charAt(begin1) == s3.charAt(begin3) &&
isInterleave(s1, begin1 + 1, end1, s2, begin2, end2,
s3, begin3 + 1, end3)) ||
(begin2 < end2 && s2.charAt(begin2) == s3.charAt(begin3) &&
isInterleave(s1, begin1, end1,
s2, begin2 + 1, end2, s3, begin3 + 1, end3));
}
}
// Interleaving String
// 递归,会超时,仅用来帮助理解
class Solution {
public:
bool isInterleave(const string& s1, const string& s2, const string& s3) {
if (s3.length() != s1.length() + s2.length())
return false;
return isInterleave(begin(s1), end(s1), begin(s2), end(s2),
begin(s3), end(s3));
}
template<typename InIt>
bool isInterleave(InIt first1, InIt last1, InIt first2, InIt last2,
InIt first3, InIt last3) {
if (first3 == last3)
return first1 == last1 && first2 == last2;
return (*first1 == *first3
&& isInterleave(next(first1), last1, first2, last2,
next(first3), last3))
|| (*first2 == *first3
&& isInterleave(first1, last1, next(first2), last2,
next(first3), last3));
}
};
动规
// Interleaving String
// 二维动规,时间复杂度O(n^2),空间复杂度O(n^2)
public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if (s3.length() != s1.length() + s2.length())
return false;
boolean[][] f = new boolean[s1.length() + 1][s2.length() + 1];
for (int i = 0; i < s1.length() + 1; ++i)
Arrays.fill(f[i], true);
for (int i = 1; i <= s1.length(); ++i)
f[i][0] = f[i - 1][0] && s1.charAt(i - 1) == s3.charAt(i - 1);
for (int i = 1; i <= s2.length(); ++i)
f[0][i] = f[0][i - 1] && s2.charAt(i - 1) == s3.charAt(i - 1);
for (int i = 1; i <= s1.length(); ++i)
for (int j = 1; j <= s2.length(); ++j)
f[i][j] = (s1.charAt(i - 1) == s3.charAt(i + j - 1) && f[i - 1][j])
|| (s2.charAt(j - 1) == s3.charAt(i + j - 1) && f[i][j - 1]);
return f[s1.length()][s2.length()];
}
}
// Interleaving String
// 二维动规,时间复杂度O(n^2),空间复杂度O(n^2)
class Solution {
public:
bool isInterleave(const string& s1, const string& s2, const string& s3) {
if (s3.length() != s1.length() + s2.length())
return false;
vector<vector<bool>> f(s1.length() + 1,
vector<bool>(s2.length() + 1, true));
for (size_t i = 1; i <= s1.length(); ++i)
f[i][0] = f[i - 1][0] && s1[i - 1] == s3[i - 1];
for (size_t i = 1; i <= s2.length(); ++i)
f[0][i] = f[0][i - 1] && s2[i - 1] == s3[i - 1];
for (size_t i = 1; i <= s1.length(); ++i)
for (size_t j = 1; j <= s2.length(); ++j)
f[i][j] = (s1[i - 1] == s3[i + j - 1] && f[i - 1][j])
|| (s2[j - 1] == s3[i + j - 1] && f[i][j - 1]);
return f[s1.length()][s2.length()];
}
};
动规+滚动数组
// Interleaving String
// 二维动规+滚动数组,时间复杂度O(n^2),空间复杂度O(n)
public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if (s1.length() + s2.length() != s3.length())
return false;
if (s1.length() < s2.length())
return isInterleave(s2, s1, s3);
boolean[] f = new boolean[s2.length() + 1];
Arrays.fill(f, true);
for (int i = 1; i <= s2.length(); ++i)
f[i] = s2.charAt(i - 1) == s3.charAt(i - 1) && f[i - 1];
for (int i = 1; i <= s1.length(); ++i) {
f[0] = s1.charAt(i - 1) == s3.charAt(i - 1) && f[0];
for (int j = 1; j <= s2.length(); ++j)
f[j] = (s1.charAt(i - 1) == s3.charAt(i + j - 1) && f[j])
|| (s2.charAt(j - 1) == s3.charAt(i + j - 1) && f[j - 1]);
}
return f[s2.length()];
}
}
// Interleaving String
// 二维动规+滚动数组,时间复杂度O(n^2),空间复杂度O(n)
class Solution {
public:
bool isInterleave(const string& s1, const string& s2, const string& s3) {
if (s1.length() + s2.length() != s3.length())
return false;
if (s1.length() < s2.length())
return isInterleave(s2, s1, s3);
vector<bool> f(s2.length() + 1, true);
for (size_t i = 1; i <= s2.length(); ++i)
f[i] = s2[i - 1] == s3[i - 1] && f[i - 1];
for (size_t i = 1; i <= s1.length(); ++i) {
f[0] = s1[i - 1] == s3[i - 1] && f[0];
for (size_t j = 1; j <= s2.length(); ++j)
f[j] = (s1[i - 1] == s3[i + j - 1] && f[j])
|| (s2[j - 1] == s3[i + j - 1] && f[j - 1]);
}
return f[s2.length()];
}
};