87. Scramble String
难度:Hard
刷题内容
原题连接
- https://leetcode.com/problems/scramble-string/
内容描述
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
˼·1 **- ʱ¼ä¸´ÔÓ¶È: O(2^n)*- ¿Õ¼ä¸´ÔÓ¶È: O(1)***
¸Õ¿ªÊ¼µÄʱºòÌâÄ¿µÄÒâ˼ûÀí½âÇå³þ£¬ÒÔΪÊÇ°Ñ×Ö·û´®¶Ô°ë£¬Æäʵ×Ö·û´®ÓкܶàµÄ¶þ²æÊ÷±íʾ·½·¨¡£µÚÒ»ÖÖ·½·¨¾ÍÊǵݹéµÄ·½·¨¡£Ã¿´ÎµÝ¹éʱ±È½Ïiµ½lenµÄÖ®¼äµÄ×Ó´®ÊÇ·ñºÍs2ÖеÄ×Ó´®µÄÏàͬ£¬ÕâÀïÓÐÁ½ÖֱȽϷ½Ê½£¬s2¿ªÊ¼ºÍ½áβÁ½ÖֱȽϷ½Ê½£¬ÓÐÒ»ÖÖÂú×ã¼´¿É£¬ÕâÀïÎÒÃÇÖ»ÒªÅжÏÁ½¸ö×Ó´®µÄ×ÖĸÊÇ·ñÏàͬ¡£Ïàͬ¾Í½øÐÐÏÂÒ»´ÎµÝ¹é¡£
class Solution {
public:
bool isScramble(string s1, string s2) {
if(s1==s2)
return true;
int len = s1.length();
int count[26] = {0};
for(int i=0; i<len; i++)
{
count[s1[i]-'a']++;
count[s2[i]-'a']--;
}
for(int i=0; i<26; i++)
{
if(count[i]!=0)
return false;
}
for(int i=1; i<=len-1; i++)
{
if(isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i)))
return true;
if(isScramble(s1.substr(0,i), s2.substr(len-i)) && isScramble(s1.substr(i), s2.substr(0,len-i)))
return true;
}
return false;
}
};
˼·2 **- ʱ¼ä¸´ÔÓ¶È: O(n^4)****- ¿Õ¼ä¸´ÔÓ¶È: O(n^3
µÚ¶þÖֵķ½·¨Ê¹ÓÃDP£¬¶¨ÒåÊý×édp[i][j][t],±íʾs1[i]ºÍs2[j]¿ªÊ¼µÄ³¤¶ÈΪtµÄ×Ö·ûÊÇ·ñΪscramble string¡£
class Solution {
public:
bool isScramble(string s1, string s2) {
int len = s1.length();
int dp[len][len][len + 1] = {0};
for(int i = len - 1;i >= 0;--i)
for(int j = len - 1;j >= 0;--j)
for(int t = 1;t <= min(len - i,len - j);++t)
if(t == 1)
dp[i][j][t] = (s1[i] == s2[j]);
else
{
int k = 1;
for(;k < t;++k)
if((dp[i][j][k] && dp[i + k][j + k][t - k]) || (dp[i][j + t - k][k] && dp[i + k][j][t - k]))
break;
dp[i][j][t] = k == t ? 0 : 1;
}
return dp[0][0][len];
}
};
