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.
使用DP, 过程如下图所示：
|
S2 |
|||||||
|
S1 |
aadbbcbcac |
0 “” |
1 d |
2 db |
3 dbb |
4 dbbc |
5 dbbca |
|
0 “” |
T |
F(d!=a) |
F |
F |
F |
F |
|
|
1 a |
T(a==a) |
F(aa) |
F |
F |
F |
F |
|
|
2 aa |
T |
T(aad) |
T(aadb) |
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|
|
|
|
3 aab |
F(aab!=aad) |
T(aadb) |
T(aadbb) |
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|
|
|
|
4 aabc |
F |
F(aadbb) |
T(aadbbc) |
|
|
|
|
|
5 aabcc |
F |
F(aadbbc) |
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|
|
|
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发现某一格dp[i][j]为true只有当其上面或左边为true才行。且需要新加入的字母与s3新添加的字母一致，True状态才能延续。
代码：
1 public boolean isInterleave(String s1, String s2, String s3) { 2 int l1=s1.length(); 3 int l2=s2.length(); 4 int l3=s3.length(); 5 if(l1+l2!=l3) 6 return false; 7 boolean[][] dp = new boolean[l1+1][l2+1]; 8 for(int i=0;i<=l1;i++) { 9 for(int j=0;j<=l2;j++) { 10 if(i==0 && j==0) 11 dp[i][j]=true; 12 else if(i==0) 13 dp[i][j] = dp[i][j-1] && (s2.charAt(j-1)==s3.charAt(i+j-1)); 14 else if(j==0) 15 dp[i][j] = dp[i-1][j] && (s1.charAt(i-1)==s3.charAt(i+j-1)); 16 else 17 dp[i][j] = (dp[i][j-1] && (s2.charAt(j-1)==s3.charAt(i+j-1))) || (dp[i-1][j] && (s1.charAt(i-1)==s3.charAt(i+j-1))); 18 } 19 } 20 return dp[l1][l2]; 21 }
简化为一维数组：
1 public boolean isInterleave(String s1, String s2, String s3) { 2 int l1 = s1.length(); 3 int l2 = s2.length(); 4 int l3 = s3.length(); 5 if(l3!=(l1+l2)) 6 return false; 7 boolean[] dp = new boolean[l2+1]; 8 dp[0] = true; 9 for(int i=1;i<dp.length;i++) 10 { 11 dp[i] = dp[i-1] && (s2.charAt(i-1)==s3.charAt(i-1)); 12 } 13 for(int i=1;i<=l1;i++) 14 { 15 for(int j=0;j<dp.length;j++) 16 { 17 if(j==0) 18 dp[j] = dp[j] && (s1.charAt(i-1)==s3.charAt(i-1)); 19 else 20 dp[j] = (dp[j] && (s1.charAt(i-1)==s3.charAt(i+j-1))) || (dp[j-1] && (s2.charAt(j-1)==s3.charAt(i+j-1))); 21 } 22 } 23 24 return dp[l2]; 25 }
