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.
递归的来搞。比较粗暴。
class Solution {
public:
bool isScramble(string s1, string s2) {
int n = s1.size();
if (!isSameChar(s1, s2)) return false;
if (s1 == s2) return true;
for (int i = 1; i < n; i++) {
string s11 = s1.substr(0, i);
string s12 = s1.substr(i);
string s21 = s2.substr(0, i);
string s22 = s2.substr(i);
if (isScramble(s11, s21) && isScramble(s12, s22)) return true;
s21 = s2.substr(0, n - i);
s22 = s2.substr(n - i);
if (isScramble(s11, s22) && isScramble(s12,s21)) return true;
}
return false;
}
bool isSameChar(string a, string b) {
int c[256] = {0};
for (int i = 0; i < a.size(); i++) {
c[a[i]]++;
}
for (int i = 0; i < b.size(); i++) {
if (--c[b[i]] < 0) return false;
}
return true;
}
};