Leetcode | | 97, interleaving String

Problem:

Given s1, S2, S3, find whether S3 are 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.

Hide TagsDynamic Programming StringTest instructions: Can be interspersed with a third string in two string order

Thinking:

(1) First think of recursion, commit timeout.

(2) The correct solution of this problem is two-dimensional DP, state transfer equation:

Very good problem solving: http://www.cnblogs.com/remlostime/archive/2012/11/25/2787297.html

F[I][J] = (F[i][j-1] && s2[j-1] = = s3[i+j-1]) | | (F[i-1][j] && s1[i-1] = = s3[i+j-1]);

Only S3 's i+i-1 character and S1 's i-1 character or S2 's j-1 character typeface can be valid at the same time. Also refer to the previous results.

Code

Dp:

Class Solution {private:    bool f[1000][1000];p ublic:    bool Isinterleave (string s1, String s2, string s3) {        if ( S1.size () + s2.size ()! = S3.size ())            return false;                    F[0][0] = true;        for (int i = 1; I <= s1.size (); i++)  //f[i][0] Initialize            f[i][0] = f[i-1][0] && (s3[i-1] = = s1[i-1]);                    for (int j = 1; J <= S2.size (), j + +)  //f[0][j] Initialize            f[0][j] = f[0][j-1] && (s3[j-1] = = s2[j-1]);                    for (int i = 1; I <= s1.size (), i++) for            (int j = 1, J <= S2.size (); j + +)                f[i][j] = (F[i][j-1] && s2 [J-1] = = s3[i+j-1]) | | (F[i-1][j] && s1[i-1] = = s3[i+j-1]);                        Return F[s1.size ()][s2.size ()];}    ;

Recursion: Timeout

Class Solution {public:    bool Isinterleave (string s1, String s2, string s3) {        s1.push_back (' 0 ');        S2.push_back (' 0 ');        S3.push_back (' 0 ');        BOOL Flag=false;        Check (s1,0,s2,0,s3,0,flag);        return flag;    } Protected:    void Check (string s1,int i,string s2,int j,string s3,int K, bool &flag)    {        if (flag)            return;        if (s1[i]== ' 0 ' &&s2[j]== ' 0 ' &&s3[k]== ' 0 ')        {            flag=true;            return;        }        if (s1[i]== ' 0 ' &&s2[j]== ' 0 ' &&s3[k]!= ' 0 ')            return;        if ((s1[i]!= ' 0 ' &&s2[j]== ' 0 ' &&s3[k]== ' 0 ') | | (s1[i]== ' 0 ' &&s2[j]!= ' 0 ' &&s3[k]== ' 0 '))            return;        if (S1[i]!=s3[k]&&s2[j]!=s3[k])            return;        if (S1[i]==s3[k])            check (s1,i+1,s2,j,s3,k+1,flag);        if (S2[j]==s3[k])            check (S1,i,s2,j+1,s3,k+1,flag);      };

Leetcode | | 97, interleaving String