Leetcode 87 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.

判断一个字符串通过交换左右区间及其子区间的操作能否变成目标字符串。

DFS搜一下，至于有人说的DP，本质上和搜索是一样的，只不过是搜索的迭代过程罢了！

状态为dp[i][j][len]，其中i是s1的起始字符，j是s2的起始字符，而n是当前的字符串长度，整个状态表示以i和j分别为s1和s2起点的长度为len的字符串是否为Scramble String。

转移方程为：dp[i][j][len] ||= (dp[i][j][k]&&dp[i+k][j+k][len-k] || dp[i][j+len-k][k]&&dp[i+k][j][len-k])。

最开始的搜索贴上来

class Solution { public: bool isScramble(string s1, string s2) { if(s1==s2) return true; int cnt[256]={0}; for(int i=0;i<s1.size();i++) { cnt[s1[i]]++; cnt[s2[i]]--; } for(char i='a';i<='z';i++) if(cnt[i]!=0) return false; for(int i=1;i<s1.size();i++) if((isScramble(s1.substr(0,i),s2.substr(0,i)) && isScramble(s1.substr(i),s2.substr(i))) || (isScramble(s1.substr(0,i),s2.substr(s1.size()-i)) && isScramble(s1.substr(i),s2.substr(0,s1.size()-i)))) return true; return false; } };