一开始拿到这个题的时候没什么想法，浆糊了之后立马百度之，才有了思路。
简单的说，就是s1和s2是scramble的话，那么必然存在一个在s1上的长度l1，将s1分成s11和s12两段，同样有s21和s22。
那么要么s11和s21是scramble的并且s12和s22是scramble的；
要么s11和s22是scramble的并且s12和s21是scramble的。
先用递归写了一个算法，但是大集合不过，然后用三维动态规划才搞定。
题目描述Scramble String
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.
代码：三维动态规划，48ms过大集合
bool isScrambleDP(string s1, string s2) { if(s1.size() != s2.size()) return false; if(s1.size() == 0) return true; if(s1 == s2) return true; bool ***iss = NULL;//iss[len][startIndexAtS1][startIndexAtS2] iss = new bool**[s1.size()]; for(int len = 1;len <= s1.size(); ++len) { //size at this level int levelSize = s1.size() - len + 1; int levelIndex = len - 1; iss[levelIndex] = new bool*[levelSize]; for(int indexS1 = 0;indexS1 < levelSize; ++ indexS1) { iss[levelIndex][indexS1] = new bool[levelSize]; for(int is2 = 0; is2 < levelSize; ++is2) { if(len == 1) { iss[levelIndex][indexS1][is2] = (s1[indexS1] == s2[is2]); } else { iss[levelIndex][indexS1][is2] = false; for(int seglen1 = 1; seglen1 < len; ++seglen1) { int seglen2 = len - seglen1; int sli1 = seglen1 - 1; int sli2 = seglen2 - 1; if(iss[sli1][indexS1][is2] && iss[sli2][indexS1 + seglen1][is2 + seglen1]) { iss[levelIndex][indexS1][is2] = true; break; } if(iss[sli1][indexS1][is2 + seglen2] && iss[sli2][indexS1 + seglen1][is2]) { iss[levelIndex][indexS1][is2] = true; break; } } } } } } return iss[s1.size() - 1][0][0]; }
递归代码，小集合12ms过，大集合过不了，因为时间复杂度是O(3n)
class Solution { public: bool isScramble(string s1, string s2) { // Start typing your C/C++ solution below // DO NOT write int main() function if(s1.size() != s2.size()) return false; if(s1 == s2) return true; for(int isep = 1; isep < s1.size(); ++ isep) { //seporate s1 as [0,isep - 1],[isep, s1.size() - 1] string seg11 = s1.substr(0,isep); string seg12 = s1.substr(isep); {//see if forward order is ok string seg21 = s2.substr(0,isep); string seg22 = s2.substr(isep); if(isScramble(seg11,seg21) && isScramble(seg12,seg22)) return true; } {//see if reverse order is ok string seg21 = s2.substr(s2.size() - isep); string seg22 = s2.substr(0,s2.size() - isep); if(isScramble(seg11,seg21) && isScramble(seg12,seg22)) return true; } } return false; } };
Recursive code rewrite at 2013-2-5, 48ms pass large set
class Solution { public: bool isScramble(string s1, string s2) { // Start typing your C/C++ solution below // DO NOT write int main() function if(s1.size() != s2.size()) return false; if(s1 == s2) return true; string ts1 = s1,ts2 = s2; sort(ts1.begin(), ts1.end()); sort(ts2.begin(), ts2.end()); if(ts1 != ts2) return false; for(int isep = 1; isep < s1.size(); ++ isep) { //seporate s1 as [0,isep - 1],[isep, s1.size() - 1] string seg11 = s1.substr(0,isep); string seg12 = s1.substr(isep); {//see if forward order is ok string seg21 = s2.substr(0,isep); string seg22 = s2.substr(isep); if(isScramble(seg11,seg21) && isScramble(seg12,seg22)) return true; } {//see if reverse order is ok string seg21 = s2.substr(s2.size() - isep); string seg22 = s2.substr(0,s2.size() - isep); if(isScramble(seg11,seg21) && isScramble(seg12,seg22)) return true; } } return false; } };