题目：

题解：

func isScramble(s1, s2 string) bool {n := len(s1)dp := make([][][]int8, n)for i := range dp {dp[i] = make([][]int8, n)for j := range dp[i] {dp[i][j] = make([]int8, n+1)for k := range dp[i][j] {dp[i][j][k] = -1}}}// 第一个字符串从 i1 开始，第二个字符串从 i2 开始，子串的长度为 length// 和谐返回 1，不和谐返回 0var dfs func(i1, i2, length int) int8dfs = func(i1, i2, length int) (res int8) {d := &dp[i1][i2][length]if *d != -1 {return *d}defer func() { *d = res }()// 判断两个子串是否相等x, y := s1[i1:i1+length], s2[i2:i2+length]if x == y {return 1}// 判断是否存在字符 c 在两个子串中出现的次数不同freq := [26]int{}for i, ch := range x {freq[ch-'a']++freq[y[i]-'a']--}for _, f := range freq[:] {if f != 0 {return 0}}// 枚举分割位置for i := 1; i < length; i++ {// 不交换的情况if dfs(i1, i2, i) == 1 && dfs(i1+i, i2+i, length-i) == 1 {return 1}// 交换的情况if dfs(i1, i2+length-i, i) == 1 && dfs(i1+i, i2, length-i) == 1 {return 1}}return 0}return dfs(0, 0, n) == 1
}