[LeetCode] Scramble String 解题报告
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.
» Solve this problem
[解题思路]
首先想到的是递归，简单明了，对两个string进行partition，然后比较四个字符串段。但是递归的话，这个时间复杂度比较高。然后想到能否DP，但是即使用DP的话，也要O(n^3)。想想算了，还是在递归里做些剪枝，这样就可以避免冗余计算：
- 对于每两个要比较的partition，统计他们字符出现次数，如果不相等返回。
但是我没办法证明他这个算法的正确性。感觉这道题应该有O(n)的解法，一时想不出来。
1: bool isScramble(string s1, string s2) {
2: // Start typing your C/C++ solution below
3: // DO NOT write int main() function
4: if(s1.size() != s2.size()) return false;
5: int A[26];
6: memset(A,0,26*sizeof(A[0]));
7: for(int i =0;i<s1.size(); i++)
8: {
9: A[s1[i]-'a']++;
10: }
11: for(int i =0;i<s2.size(); i++)
12: {
13: A[s2[i]-'a']--;
14: }
15: for(int i =0;i<26; i++)
16: {
17: if(A[i] !=0)
18: return false;
19: }
20: if(s1.size() ==1 && s2.size() ==1) return true;
21: for(int i =1; i< s1.size(); i++)
22: {
23: bool result= isScramble(s1.substr(0, i), s2.substr(0, i))
24: && isScramble(s1.substr(i, s1.size()-i), s2.substr(i, s1.size()-i));
25: result = result || (isScramble(s1.substr(0, i), s2.substr(s2.size() - i, i))
26: && isScramble(s1.substr(i, s1.size()-i), s2.substr(0, s1.size()-i)));
27: if(result) return true;
28: }
29: return false;
30: }