87. Scramble String
题目
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.
解析
题意在于判断一个字符串是否为另一个字符串“乱序”得到，这种乱序采用的方式是将一个字符串从某个位置“割开”，形成两个子串，然后对两个子串进行同样的“割开”操作，直到到达叶子节点，无法再分割。然后对非叶子节点的左右孩子节点进行交换，最后重新从左至右组合形成新的字符串，由于这个过程中存在字符位置的变化，因此，原来的字符串顺序可能会被打乱，当然也可能没有（同一个非叶子节点的左右孩子交换0次或偶数次，就无变化）。需要注意的点：
1、原字符串每次被割开的位置并不确定可能为[1,s.size()-1]，所以必然需要遍历所有可能割开的位置；
2、原字符串从第i个位置被割开（i在区间[1,s.size()-1]），形成的两个子串s.substr(0,i)和s.substr(i,s.size()-i)，如果这两个子串不全为空，则它们的母串（这里指原字符串）就是所谓的非叶子节点，这两个子串可以左右交换（按照二叉树的展开方式）；对于两个子串，可继续割裂，直到形成叶子节点。
// 87. Scramble String
class Solution_87 {
public:
bool isScramble(string s1, string s2) {
if (s1==s2)
{
return true;
}
if (s1.size()!=s2.size())
{
return false;
}
vector<int> hash(26,0);
for (int i = 0; i < s1.size();i++)
{
hash[s1[i] - 'a']++;
hash[s2[i] - 'a']--;
}
for (int i = 0; i < 26;i++) //递归剪枝
{
if (hash[i]!=0)
{
return false;
}
}
bool res = false;
for (int i = 1; i < s1.size();i++) //遍历所有可能割开的位置, 切割的长度
{
res = res || (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i, s1.size() - i), s2.substr(i, s1.size() - i))); //长度要一致
res = res || (isScramble(s1.substr(0, i), s2.substr(s1.size() - i)) && isScramble(s1.substr(i),s2.substr(0, s1.size()-i)));
}
return res;
}
};