# 19.2.23 [LeetCode 87] Scramble String

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

## 题解

 1 class Solution {
2 public:
3     bool equal(vector<int>&a, vector<int>&b) {
4         for (int i = 0; i < a.size(); i++)
5             if (a[i] != b[i])
6                 return false;
7         return true;
8     }
9     bool isScramble(string s1, string s2) {
10         vector<int>c1(256, 0), c2(256, 0);
11         int l = s1.length();
12         if (l <= 3) {
13             for (int i = 0; i < l; i++)
14                 c1[s1[i]]++, c2[s2[i]]++;
15             if (equal(c1, c2))return true;
16             else return false;
17         }
18         for (int i = 0; i < l - 1; i++) {
19             c1[s1[i]]++, c2[s2[i]]++;
20             if (equal(c1, c2) && isScramble(s1.substr(0, i + 1), s2.substr(0, i + 1)) && isScramble(s1.substr(i + 1), s2.substr(i + 1)))
21                 return true;
22         }
23         c1 = vector<int>(256, 0), c2 = vector<int>(256, 0);
24         for (int i = 0; i < l - 1; i++) {
25             c1[s1[i]]++, c2[s2[l-1-i]]++;
26             if (equal(c1, c2) && isScramble(s1.substr(0, i + 1), s2.substr(l-1-i)) && isScramble(s1.substr(i + 1), s2.substr(0,l-i-1)))
27                 return true;
28         }
29         return false;
30     }
31 };
View Code

 1 class Solution {
2 public:
3     bool equal(vector<int>&a, vector<int>&b) {
4         for (int i = 0; i < a.size(); i++)
5             if (a[i] != b[i])
6                 return false;
7         return true;
8     }
9     bool isScramble(string s1, string s2) {
10         vector<int>c1(26, 0), c2(26, 0);
11         int l = s1.length();
12         if (l <= 3) {
13             for (int i = 0; i < l; i++)
14                 c1[s1[i]-'a']++, c2[s2[i]-'a']++;
15             if (equal(c1, c2))return true;
16             else return false;
17         }
18         for (int i = 0; i < l - 1; i++) {
19             c1[s1[i]-'a']++, c2[s2[i]-'a']++;
20             if (equal(c1, c2) && isScramble(s1.substr(0, i + 1), s2.substr(0, i + 1)) && isScramble(s1.substr(i + 1), s2.substr(i + 1)))
21                 return true;
22         }
23         c1 = vector<int>(256, 0), c2 = vector<int>(256, 0);
24         for (int i = 0; i < l - 1; i++) {
25             c1[s1[i]-'a']++, c2[s2[l-1-i]-'a']++;
26             if (equal(c1, c2) && isScramble(s1.substr(0, i + 1), s2.substr(l-1-i)) && isScramble(s1.substr(i + 1), s2.substr(0,l-i-1)))
27                 return true;
28         }
29         return false;
30     }
31 };
View Code

 1 class Solution {
2 public:
3     bool isScramble(string s1, string s2) {
4         int l = s1.length();
5         if (s1 == s2)return true;
6         string str1 = s1, str2 = s2;
7         sort(str1.begin(), str1.end());
8         sort(str2.begin(), str2.end());
9         if (str1 != str2)return false;
10         for (int i = 0; i < l - 1; i++) {
11             string p1 = s1.substr(0, i + 1), p2 = s2.substr(0, i + 1), p3 = s1.substr(i + 1), p4 = s2.substr(i + 1);
12             if (isScramble(p1, p2) && isScramble(p3, p4))return true;
13             p2 = s2.substr(l - 1 - i), p4 = s2.substr(0, l - 1 - i);
14             if (isScramble(p1, p2) && isScramble(p3, p4))return true;
15         }
16         return false;
17     }
18 };
View Code

 1 // DP
2 class Solution {
3 public:
4     bool isScramble(string s1, string s2) {
5         if (s1.size() != s2.size()) return false;
6         if (s1 == s2) return true;
7         int n = s1.size();
8         vector<vector<vector<bool> > > dp (n, vector<vector<bool> >(n, vector<bool>(n + 1, false)));
9         for (int i = 0; i < n; ++i) {
10             for (int j = 0; j < n; ++j) {
11                 dp[i][j][1] = s1[i] == s2[j];
12             }
13         }
14         for (int len = 2; len <= n; ++len) {
15             for (int i = 0; i <= n - len; ++i) {
16                 for (int j = 0; j <= n - len; ++j) {
17                     for (int k = 1; k < len; ++k) {
18                         if ((dp[i][j][k] && dp[i + k][j + k][len - k]) || (dp[i + k][j][len - k] && dp[i][j + len - k][k])) {
19                             dp[i][j][len] = true;
20                         }
21                     }
22                 }
23             }
24         }
25         return dp[0][0][n];
26     }
27 };
View Code

posted @ 2019-02-23 10:57  TobicYAL  阅读(183)  评论(0编辑  收藏  举报