87. Scramble String
problem description
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.
Example 1:
Input: s1 = "great", s2 = "rgeat"
Output: true
Example 2:
Input: s1 = "abcde", s2 = "caebd"
Output: false
algorithm thought
判断是否Scramble,可以用递归的方式进行。如果一个字符串的左边和另一个字符串的左边是scramble,右边和右边是scramble。或者是,左右,右左是scramble字符串的话。那么这两个字符串就是scramble字符串。可以通过剪枝加快判断,首先判断两个字符串是否包含同样的字符,两个字符串中包含的字符不一样,就可以直接返回。虽然这里用了O(n)的时间,但是能剪去很多分支,是值得的。
code
class Solution {
public:
bool isScramble(string s1, string s2) {
// cout<<s1<<':'<<s2<<' ';
if(s1==s2)
return true;
int count[26]={0};
int n=s1.size();
for(int i=0;i<n;++i){
count[int(s1[i]-'a')]++;
count[int(s2[i]-'a')]--;
}
for(int i=0;i<26;++i)
if(count[i]!=0)
return false;
for(int i=1;i<n;++i){
if(isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i),s2.substr(i)))
return true;
if(isScramble(s1.substr(n-i),s2.substr(0,i))&&isScramble(s1.substr(0,n-i),s2.substr(i,n-i)))
return true;
}
return false;
}
};
algorithm analysis
每个函数正常情况下都能分出4个分支,每个分支中时间复杂度是线性的。最后时间复杂度不好定义。但是整个运行时间复杂度还是挺高的
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