// Source : https://oj.leetcode.com/problems/scramble-string/ // Author : Hao Chen // Date : 2014-10-09 /********************************************************************************** * * 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. * * **********************************************************************************/ #include #include #include #include #include #include using namespace std; // The recursive way is quite simple. // 1) break the string to two parts: // s1[0..j] s1[j+1..n] // s2[0..j] s2[j+1..n] // 2) then // isScramble(s1[0..j], s2[0..j]) && isScramble(s1[j+1..n], s2[j+1..n]) // OR // isScramble(s1[0..j], s2[j+1, n]) && isScramble(s1[j+1..n], s2[0..j]) bool isScramble_recursion(string s1, string s2) { if (s1.size()!= s2.size() || s1.size()==0 || s2.size()==0) { return false; } if (s1 == s2){ return true; } string ss1 = s1; string ss2 = s2; sort(ss1.begin(), ss1.end()); sort(ss2.begin(), ss2.end()); if (ss1 != ss2 ) { return false; } for (int i=1; i > > dp(len+1, vector< vector >(len, vector(len) ) ); // ignor the k=0, just for readable code. // initialization k=1 for (int i=0; i2){ s1 = argv[1]; s2 = argv[2]; } cout << s1 << ", " << s2 << endl; cout << isScramble(s1, s2) << endl; return 0; }