"One Day together Leetcode" #87. Scramble String

One Day together Leetcode

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(i) Title

Given A string s1, we may be represent it as a binary tree by partitioning it to the non-empty substrings recursively.

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 this "rgeat" is a scrambled string of "great".

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

Rgtae
/ \
RG Tae
/ \ / \
R g Ta E
/ \
T A

We say this "Rgtae" is a scrambled string of "great".

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

(ii) Problem solving

The main idea: Judge two string is not scramble string, the topic is a bit long, need to read a long time!!!
To give a simple example of scramble string, such as AB and BA, then AB splits into a and b,ba split into B and A, which represents the scramble string.
Then whether S1 and S2 are scramble strings, it is necessary to split the S1 and S2 into two parts S11 and S12,S21 and S22, which need to be judged whether the two parts are scramble string.
Can be divided into two parts: (S11 and S21,S12 and S22) as well as (S11 and S22,S12 and S21).
So the code is better written!
The dynamic programming problem is obvious. But I do the problem in the process of optimization is still not done, resulting in timeouts.

classSolution { Public:BOOLIsscramble (stringS1,stringS2) {intLen1 = S1.length ();intLen2 = S2.length ();if(Len1! = len2)return false;//Length not the same        if(S1 = = s2)return true;//substring equal        //Determine if two substrings contain the same characters        //Did not add this step at first, resulting in a time-out         vector<int>Count -,0); for(inti =0; i < len1; i++) {count[s1[i]-' A ']++; count[s2[i]-' A ']--; } for(inti =0; I < -; i++) {if(count[i]!=0)return false; }//Recursive         for(inti =1; i < len1; i++) {stringSubs1_1 = S1.substr (0, i);stringSubs1_2 = S1.substr (i);stringSubs2_1 = S2.substr (0, i);stringSubs2_2 = S2.substr (i);BOOLIs1 = (subs1_1==subs2_1?true: Isscramble (Subs1_1, Subs2_1)) && (subs1_2==subs2_2?true: Isscramble (Subs1_2, subs2_2));//The recursion is also optimized here, and the equivalent is not recursiveSubs2_1 = S2.substr (len2-i, i); Subs2_2 = S2.substr (0, len2-i);BOOLIs2 = (subs1_1==subs2_1?true: Isscramble (Subs1_1, Subs2_1)) && (subs1_2==subs2_2?true: Isscramble (Subs1_2, subs2_2));if(Is1 | | is2)return true;//Two part of scramble string S1 and S2 are scramble string}return false; }};

"One Day together Leetcode" #87. Scramble String