LeetCode: Scramble String

这题看了网上答案

 1 class Solution {

 2 public:

 3     bool isScramble(string s1, string s2) {

 4         // Start typing your C/C++ solution below

 5         // DO NOT write int main() function

 6         if (s1.size() != s2.size()) return false;

 7         map<char, int> S1, S2;

 8         for (int i = 0; i < s1.size(); i++) S1[s1[i]]++;

 9         for (int i = 0; i < s2.size(); i++) S2[s2[i]]++;

10         if (S1 != S2) return false;

11         if (s1.size() == 1) return true;

12         for (int i = 1; i < s1.size(); i++) {

13             bool flag = isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i, s1.size()-i), s2.substr(i, s2.size()-i));

14             flag = flag || (isScramble(s1.substr(0, i), s2.substr(s2.size()-i, i)) && isScramble(s1.substr(i, s1.size()-i), s2.substr(0, s2.size()-i)));

15             if (flag) return true;

16         }

17         return false;

18     }

19 };

 三维DP更牛逼！

 1 class Solution {

 2 public:

 3     bool isScramble(string s1, string s2) {

 4         if (s1.size() != s2.size()) return false;

 5         int n = s1.size();

 6         vector<vector<vector<bool> > > f(n, vector<vector<bool> >(n, vector<bool>(n)));

 7         for (int i = 0; i < n; ++i) {

 8             for (int j = 0; j < n; ++j) {

 9                 f[i][j][0] = (s1[i] == s2[j]);

10             }

11         }

12         for (int l = 1; l < n; ++l) {

13             for (int i = 0; i+l < n; ++i) {

14                 for (int j = 0; j+l < n; ++j) {

15                     for (int k = 0; k < l; ++k) {

16                         f[i][j][l] = f[i][j][l] || (f[i][j][k] && f[i+k+1][j+k+1][l-1-k] || f[i][j+l-k][k] && f[i+k+1][j][l-1-k]);

17                     }

18                 }

19             }

20         }

21         return f[0][0][n-1];

22     }

23 };

 C#

 1 public class Solution {

 2     public bool IsScramble(string s1, string s2) {

 3         if (s1.Length != s2.Length) return false;

 4         Dictionary<char, int> S1 = new Dictionary<char, int>();

 5         Dictionary<char, int> S2 = new Dictionary<char, int>();

 6         for (int i = 0; i < s1.Length; i++) {

 7             if (S1.ContainsKey(s1[i])) S1[s1[i]]++;

 8             else S1.Add(s1[i], 1);

 9         }

10         for (int i = 0; i < s2.Length; i++) {

11             if (S2.ContainsKey(s2[i])) S2[s2[i]]++;

12             else S2.Add(s2[i], 1);

13         }

14         foreach (var v in S1) {

15             if (!S2.ContainsKey(v.Key) || v.Value != S2[v.Key]) return false;

16         }

17         if (s1.Length == 1) return true;

18         for (int i = 1; i < s1.Length; i++) {

19             bool flag = IsScramble(s1.Substring(0, i), s2.Substring(0, i)) && IsScramble(s1.Substring(i, s1.Length-i), s2.Substring(i, s2.Length-i));

20             flag = flag || (IsScramble(s1.Substring(0, i), s2.Substring(s2.Length-i, i)) && IsScramble(s1.Substring(i, s1.Length-i), s2.Substring(0, s2.Length-i)));

21             if (flag) return true;

22         }

23         return false;

24     }

25 }

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