标签：des style blog color 使用 os io for
Given a string s1, we may represent it as a binary tree by partitioning it to two 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 may choose any non-leaf node and swap its two children.
For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".
rgeat
/ rg eat
/ \ / r g e at
/ a t
We say that "rgeat" is a scrambled string of "great".
Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".
rgtae
/ rg tae
/ \ / r g ta e
/ t a
We say that "rgtae" is a scrambled string of "great".
Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.
思路：枚举所有的分割点。假设s1={ s11, s12 }，则可能的匹配方式为：s2={ scramble(s11), scramble(s12) }或者s2={ scramble(s12), scramble(s11) }。这样，记s2={s21,s22}，问题被分解为求解：isScramble(s11,s21) && isScramble(s12,s22) 和 isScramble(s11,s22) && isScramble(s12,s21)。显然，子问题的规模较原问题变小了。
使用DP来求解该题，Q[step][i][j]记录以s1[i]起始长度为step+1的子串（即s1.substr(i,step+1)）和以s2[j]起始长度为step+1的子串（即s2.substr(j,step+1)）之间是否可转换。则，Q[step][i][j] = ( Q[k][i][j] && Q[step-k-1][i+k+1][j+k+1] ) || ( Q[k][i][j+step-k] && Q[step-k-1][i+k+1][j] )，其中，k=0, 1, ..., step-1，即依次取s1[i],s1[i+1],...,s1[step]作为子串s1.substr(i,step+1)的分割点。
1 class Solution { 2 public: 3 bool isScramble( string s1, string s2 ) { 4 if( s1.length() != s2.length() ) { return false; } 5 int slen = s1.length(); 6 vector<vector<vector<bool>>> Q( slen ); 7 for( int step = 0; step < slen; ++step ) { 8 Q[step].resize( slen-step, vector<bool>( slen-step, false ) ); 9 vector<vector<bool>>& Qi = Q[step]; 10 if( step == 0 ) { 11 for( size_t i = 0; i != Qi.size(); ++i ) { 12 for( size_t j = 0; j != Qi[0].size(); ++j ) { 13 if( s1[i] == s2[j] ) { Qi[i][j] = true; } 14 } 15 } 16 } else { 17 for( size_t i = 0; i != Qi.size(); ++i ) { 18 for( size_t j = 0; j != Qi[0].size(); ++j ) { 19 for( int k = 0; k < step; ++k ) 20 { 21 if( ( Q[k][i][j] && Q[step-k-1][i+k+1][j+k+1] ) 22 || ( Q[k][i][j+step-k] && Q[step-k-1][i+k+1][j] ) ) { 23 Qi[i][j] = true; 24 break; 25 } 26 } 27 } 28 } 29 } 30 } 31 return Q.back()[0][0]; 32 } 33 };
Scramble String,布布扣,bubuko.com
标签：des style blog color 使用 os io for
原文地址：http://www.cnblogs.com/moderate-fish/p/3925524.html
