LeetCode: Interleaving String

第一次dfs， 没过large， 马上想到dp， 一次过了

 1 class Solution {

 2 public:

 3     bool isInterleave(string s1, string s2, string s3) {

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

 5         // DO NOT write int main() function    

 6         if (s3.size() != s1.size()+s2.size()) return false;

 7         int m = s1.size();

 8         int n = s2.size();

 9         vector<vector<bool>> f(m+1, vector<bool>(n+1));

10         f[0][0] = true;

11         for (int i = 1; i <= m; i++) {

12             f[i][0] = s1[i-1] == s3[i-1]? f[i-1][0]:false;

13         }

14         for (int i = 1; i <= n; i++) {

15             f[0][i] = s2[i-1] == s3[i-1]? f[0][i-1]:false;

16         }

17         for (int i = 1; i <= m; i++) {

18             for (int j = 1; j <= n; j++) {

19                 f[i][j] = (f[i-1][j] && s1[i-1] == s3[i+j-1]) || (f[i][j-1] && s2[j-1] == s3[i+j-1]);

20             }

21         }

22         return f[m][n];

23     }

24 };

 C#

 1 public class Solution {

 2     public bool IsInterleave(string s1, string s2, string s3) {

 3         if (s3.Length != s1.Length + s2.Length) return false;

 4         bool[,] f = new bool[s1.Length+1, s2.Length+1];

 5         f[0, 0] = true;

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

 7             f[i, 0] = s1[i-1] == s3[i-1]? f[i-1, 0] : false;

 8         }

 9         for (int i = 1; i <= s2.Length; i++) {

10             f[0, i] = s2[i-1] == s3[i-1]? f[0, i-1] : false;

11         }

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

13             for (int j = 1; j <= s2.Length; j++) {

14                 f[i, j] = (f[i-1, j] && s1[i-1] == s3[i+j-1]) || (f[i, j-1] && s2[j-1] == s3[i+j-1]);

15             }

16         }

17         return f[s1.Length, s2.Length];

18     }

19 }

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