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.
[balabala] 暴力解决枚举出s1的全部scramble string, 存在大量重复计算;分段比较s1,s2也存在大量重复计算,比如s1=abcdef, s2=dcabef, isScramble(bcdef, cabef) 就会重复计算isScramble(cdef, abef)。使用动态规划,保存s1, s2每个位置处各种长度对应的字串的匹配情况,dp[l][i][j]表示s1[i, i+l) 和 s2[j, j+l)是否是scramble关系。
// timeout public boolean isScramble1(String s1, String s2) { if (s1 == null || s2 == null || s1.length() != s2.length()) return false; if (s1.equals(s2)) return true; int lengthS1 = s1.length(); for (int i = 1; i < lengthS1; i++) { boolean cond1 = isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i, lengthS1), s2.substring(i, lengthS1)); boolean cond2 = isScramble(s1.substring(0, i), s2.substring(lengthS1 - i, lengthS1)) && isScramble(s1.substring(i, lengthS1), s2.substring(0, lengthS1 - i)); if (cond1 || cond2) return true; } return false; } public boolean isScramble(String s1, String s2) { if (s1 == null || s2 == null || s1.length() != s2.length()) return false; if (s1.equals(s2)) return true; int length = s1.length(); boolean[][][] dp = new boolean[length + 1][length][length]; for (int l = 1; l <= length; l++) { for (int i = 0; i <= length - l; i++) { for (int j = 0; j <= length - l; j++) { if (l == 1) { dp[1][i][j] = s1.charAt(i) == s2.charAt(j); } else { for (int k = 1; k < l; k++) { dp[l][i][j] = (dp[k][i][j] && dp[l - k][i + k][j + k]) || (dp[k][i][j + l - k] && dp[l - k][i + k][j]); if (dp[l][i][j]) break; } } } } } return dp[length][0][0]; }