难度:Hard
原题连接
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
思路1 - 时间复杂度: O(2^n)- 空间复杂度: O(1)******
刚开始的时候题目的意思没理解清楚，以为是把字符串对半，其实字符串有很多的二叉树表示方法。第一种方法就是递归的方法。每次递归时比较i到len的之间的子串是否和s2中的子串的相同，这里有两种比较方式，s2开始和结尾两种比较方式，有一种满足即可，这里我们只要判断两个子串的字母是否相同。相同就进行下一次递归。
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;
}
};思路2 - 时间复杂度: O(n^4)- 空间复杂度: O(n^3
第二种的方法使用DP，定义数组dp[i][j][t],表示s1[i]和s2[j]开始的长度为t的字符是否为scramble 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];
}
};