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: }
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