[Leetcode] Scramble string String DP

Given A string S1, we may represent it as a binary tree by partitioning it to the Non-empty substrings Recursivel Y.

Below is one possible representation of S1 = "great" :

    Great   /      gr    eat/\    /  g   r  E   at           /           a   t

To scramble the string, we are choose any non-leaf node and swap it to the children.

For example, if we choose the node "gr" and swaps its-children, it produces a scrambled string "rgeat" .

    Rgeat   /      RG    eat/\    /  r   G  e   at           /           a   t

We say is "rgeat" a scrambled string of "great" .

Similarly, if we continue to swap the children of nodes "eat" "at" and, it produces a scrambled string "rgtae" .

    Rgtae   /      RG    tae/\    /  r   G  ta  e       /       t   a

We say is "rgtae" a scrambled string of "great" .

Given strings S1 and S2 of the same length, determine if S2 is a scrambled string of S1.

Hide TagsDynamic Programming String The topic began to think or is quite complex, the first idea is a binary tree search, all possible results are searched, and then if found is, can not find it is not, so write the words of speed need to consider, improve the method is depth lookup time to determine whether the number of characters in the input parameter is the same:

classSolution { Public:    BOOLIsscramble (stringS1,stringS2) {        intLen1 = S1.size (), len2 =s2.size (); if(Help_f (S1,S2)) {if(S1==S2)return true;  for(inti =1; i<len1;i++){                if(S1.substr (0, i) +s1.substr (i) ==s2)return true; }             for(intI=1; i<len1;i++){                if(Isscramble (S1.substr (0, i), S2.substr (0, i)) &&isscramble (S1.substr (i), s2.substr (i)))return true; //cout<<s1.substr (0,i) << "<<s2.substr (len2-i) <<" "<<s1.substr (len1-i) << "" <<s2.substr (0,i) <<endl;                if(Isscramble (S1.substr (0, i), S2.substr (len2-i)) &&isscramble (S1.substr (i), S2.substr (0, len2-i)))return true; }        }        return false; }    BOOLHelp_f (string&AMP;S1,string&S2) {        if(S1.size ()!=s2.size ())return false; intc[ -]={0};  for(intI=0; I<s1.size (); i++) c[s1[i]-'a'] ++;  for(intI=0; I<s2.size (); i++) {C[s2[i]-'a']--; if(c[s2[i]-'a']<0)return false; }        return true; }};

The second idea is dynamic planning, set Table[i][j][len],i J as the starting position of the string S1 s2, Len is the length to be considered, if S1 I to i + Len and S2 's J to J+len conform, true, within the length range, traverse each Location: Tab[i][j][len] |= tab[i][j][l] && tab[i+l][j+l][len-l] or Tab[i][j][len] |= tab[i][j+len-l][l] && Tab[i+l][j][len-l]

classsolution{ Public:    BOOLIsscramble (stringS1,stringS2) {        intLen1=s1.size (), len2=s2.size (); if(LEN1!=LEN2)return false; BOOLtable[ -][ -][ -]={false};  for(intI=len1-1; i>=0; i--){             for(intJ=len1-1; j>=0; j--) {table[i][j][1]= (s1[i]==S2[j]);  for(inttmplen=2; i+tmplen<=len1&&j+tmplen<=len1;tmplen++){                     for(intidx=1; idx<tmplen;idx++) {Table[i][j][tmplen]|=table[i][j][idx]&&table[i+idx][j+idx][tmplen-IDX]; Table[i][j][tmplen]|=table[i][j+tmplen-idx][idx]&&table[i+idx][j][tmplen-IDX]; }                }            }        }        returntable[0][0][LEN1]; }};

[Leetcode] Scramble string String DP