scramble-string

题目描述：

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   tTo 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   tWe 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   aWe 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.

递归AC代码：

class Solution {public:    bool isScramble(string s1, string s2) {      int len1=s1.size();      int len2=s2.size();      if (s1 == s2)         return true;      string s3=s1,s4=s2;      sort(s3.begin(),s3.end());      sort(s4.begin(),s4.end());        if(len1 != len2||s3!=s4)            return false;     for (int i = 1; i < len1; ++i){          if (isScramble(s1.substr(0, i), s2.substr(0, i))&&  isScramble(s1.substr(i), s2.substr(i)))             return true;          if (isScramble(s1.substr(0, i), s2.substr(len2 - i))&& isScramble(s1.substr(i), s2.substr(0, len2 - i)))             return true;     }      return false;     }};

动态规划AC代码：

class Solution {public:    bool isScramble(string s1, string s2) {        const int N = s1.size();        if (N != s2.size()) return false;          // f[n][i][j]，表示长度为n，起点为s1[i]和        // 起点为s2[j]两个字符串是否互为scramble        bool f[N + 1][N][N];        fill_n(&f[0][0][0], (N + 1) * N * N, false);          for (int i = 0; i < N; i++)            for (int j = 0; j < N; j++)                f[1][i][j] = s1[i] == s2[j];          for (int n = 1; n <= N; ++n) {            for (int i = 0; i + n <= N; ++i) {                for (int j = 0; j + n <= N; ++j) {                    for (int k = 1; k < n; ++k) {                        if ((f[k][i][j] && f[n - k][i + k][j + k]) ||                                (f[k][i][j + n - k] && f[n - k][i + k][j])) {                            f[n][i][j] = true;                            break;                        }                    }                }            }        }        return f[N][0][0];    }};